1 . What is the % of free SO3 in an oleum that is labelled as '104.5 % H2SO4'? (A) 10 (B) 20 (C) 40 (D) none of these 2 . 9.0 g water is added into oleum sample labelled as '112% H2SO4' then the amount of free SO3 remaining in the solution is : (A) 14.93 L at STP (B) 7.46 L at STP (C) 3.73 L at STP (D) 11.2 L at STP 3 . If excess water is added into a bottle sample labelled as '112 % H2SO4' and is reacted with 5.3 g Na2CO3, then find the volume of CO2 evolved at 1 atm pressure and 300 K temperature after the completion of the reaction : (A) 2.46 L (B) 24.6 L (C) 1.23 L (D) 12.3 L 4 . 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % of free SO3 in the sample is : (A) 74 (B) 26 (C) 20 (D) none of these Comprehension # 2 The strength of H2O2 is expressed in several ways like molarity, normality, % (w/V), volume strength, etc. The strength of \"10 V\" means 1 volume of H2O2 on decomposition gives 10 volumes of oxygen at STP or 1 litre of H2O2 gives 10 litre of O2 at STP. The decomposition of H2O2 is shown as under : H2O2 (aq) H2O () 1 O2 (g) + 2 H2O2 can acts as oxidising as well as reducing agent, as oxidizing agent H2O2 converted into H2O and as reducing agent H2O2 converted into O2, both cases it's n-factor is 2. Normality of H2O2 solution = 2 × Molarity of H2O2 solution 1 . What is the molarity of \"11.2 V\" of H2O2 ? (A) 1 M (B) 2 M (C) 5.6 M (D) 11.2 M 2 . What is the percentage strength (% w/V) of \"11.2 V\" H2O2 ? (A) 1.7 (B) 3.4 (C) 34 (D) none of these 3 . 20 mL of H2O2 solution is reacted with 80 mL of 0.05 M KMnO4 in acidic medium then what is the volume strength of H2O2 ? (A) 2.8 (B) 5.6 (C) 11.2 (D) none of these 4 . 40 g Ba(MnO4)2 (mol. wt. = 375) sample containing some inert impurities in acidic medium is completely reacted with 125 mL of \"33.6 V\" of H2O2. What is the percentage purity of the sample ? (A) 28.12 % (B) 70.31 % (C) 85 % (D) none of these Comprehension # 3 Molecular weight / Atomic weight Equivalent weight = n factor n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reaciprocal of n-factor's ratio of the reactants is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H+ /OH– furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant. Example 1 : 1. In acidic medium : KMnO (n = 5) Mn2+ 4 2. In basic medium : KMnO4 (n = 3) Mn2+ 3. In neutral medium : KMnO (n = 1) Mn6+ 4
Example 2 : FeC2O4 Fe3+ + 2CO2 Total no. of moles of e– lost by 1 mole of FeC2O4 = 1 + 1 × 2 3 n-factor of FeC2O4 = 3 1 . n-factor of Ba(MnO4)2 in acidic medium is : (A) 2 (B) 6 (C) 10 (D) none of these 2 . For the reaction, H3PO2 + NaOH NaH2PO2 + H2O What is the equivalent weight of H3PO2 ? (mol. wt. is M) (A) M (B) M/2 (C) M/3 (D) none of these 3 . For the reaction, Fe0.95O (molar mass : M) Fe2O3. What is the eq. wt. of Fe0.95O ? M M M (D) none of these (A) (B) (C) 0.8075 0.85 0.95 4 . In the reaction, xVO + yFe2O3 FeO + V2O5. What is the value of x and y respectively ? (A) 1, 1 (B) 2, 3 (C) 3, 2 (D) none of these MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 1. F 2. T 3. F 4. F 5. F Fill in the Blanks 1. decrease 2. loses 3. oxidized as well as reduced 4. oxygen has been oxidised (O2– O° ) ; chlorine has been reduced (Cl+5 Cl–1) 2 5. x = +7/3 6. +2 7. is not Match the Column 1. (A) s ; (B) p ; (C) q ; (D) r 2. (A) q ; (B) p ; (C) s ; (D) r 3. (A) s ; (B) q ; (C) r ; (D) p Assertion - Reason Questions 1. B 2. A 3. A 4. B 5. A Comprehension Based Questions Comprehension #1 : 1. (B) 2. (C) 3. (C) 4. (B) 3. (B) 4. (B) Comprehensi on #2 : 1. (A) 2. (B) 3. (C) 4. (B) Comprehensi on #3 : 1. (C) 2. (A)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Calculate the oxidation number of underlined elements in the following compounds : (a) K[ Co (C2O4)2.(NH3)2] (b) K4 P2O7 (c) Cr O2Cl2 (d) Na2[ Fe (CN)5NO+] (e) Mn3O4 (f) Ca(ClO2)2 (h) ZnO 2– (c) Fe O (g) [ Fe (NO)(H O) ]SO 4 25 2 0.93 2 . Write balanced net ionic equation for the following reactions in acidic solution. (a) S O 2– (aq) +Al (s) HS (aq) + Al3+ (aq) 46 2 (b) S O 2– (aq) + Cr O 2– (aq) S O 2– (aq) + Cr3+ (aq) 23 27 46 (c) ClO – (aq) + As S (s) Cl– (aq) + H AsO – (aq) + HSO – (aq) 3 23 24 4 (d) IO – (aq) + Re (s) ReO – (aq) + I– (aq) 3 4 (e) HSO – (aq) + As (s) + Pb O (s) PbSO (s) + H AsO – (aq) 4 4 34 4 24 (f) HNO (aq) NO – + NO (g) 2 3 3 . Write balanced net ionic equations for the following reactions in basic solution : (a) C H O 2–(aq) + ClO – (aq) CO 2– (aq) + Cl– (aq) 446 3 3 (b) Al (s) + BiONO (s) Bi (s) + NH (aq) + Al (OH) –(aq) 3 3 4 (c) HO (aq) + Cl O (aq) ClO – (aq) + O (g) 22 27 2 2 (d) Tl O (s) + NH OH (aq) TlOH (s) + N (g) 23 2 2 (e) Cu(NH ) 2+ (aq) + S O 2– (aq) SO 2– (aq) + Cu (s) + NH (aq) 34 24 3 3 (f) Mn(OH) (s) + MnO – (aq) MnO (s) 2 4 2 4 . KMnO4 oxidizes Xn+ ion to XO3–, itself changing to Mn2+ in acid medium. 2.68 × 10–3 mole of Xn+ requires 1.61 × 10–3 mole of MnO –. What is the value of n? Also calculate the atomic mass of X, if the weight 4 of 1g equivalent of XCl is 56. n 5 . In a quantitative determination of iron in an ore, an analyst converted 0.40 g, of the ore into its ferrous. This required 40.00 mL of 0.1 N solution of KMnO for titration. 4 (i) How many milliequivalents of KMnO does 40.00 mL of 0.1 N solution represent? 4 (ii) How many equivalents of iron were present in the sample of the ore taken for analysis? (iii) How many grams of iron were present in the sample? (iv) What is the percentage of iron in the ore? (v) What is the molarity of KMnO solution used? 4 (vi) How many moles of KMnO were used for titration ? (Fe = 56) 4 6 . The mixture of CuS (molar weight = M ) and Cu S (molecular weight = M ) oxidised by KMnO (molecular 12 24 weight = M3) in acidic medium, the product obtained are Cu2+, SO2. Find the equivalent weight of CuS, Cu S and KMnO respectively. 24 7. Consider the reaction H+ + IO– + I– I + H O. Find the ratio of coefficients of IO – , I– and I. 4 2 2 4 2 8 . A dilute solution of H SO is made by adding 5 mL of 3N H SO to 245 mL of water. Find the normality 24 24 and molarity of the solution. 9 . What volume at NTP of gaseous ammonia will be required to be passed into 30 cc of N – H2SO4 solution to bring down the acid strength of the latter to 0.2 N.
1 0 . A solution containing 4.2 g of KOH and Ca(OH ) is neutralized by an acid. It consums 0.1 equivalent of 2 acid, calculate the percentage composition of the sample. 1 1 . How many mL of 0.1 N HCl are required to react completely with 1 g mixture of Na CO and NaHCO 23 3 containing equimolar amounts of two? 1 2 . 0.5 g of fuming H SO (oleum) is diluted with water. The solution requires 26.7 mL of 0.4N NaOH for 24 complete neutralization. Find the % of free SO in the sample of oleum. 3 1 3 . 10 g CaCO were dissolved in 250 mL of M HCl and the solution was boiled. What volume of 2 M KOH 3 would be required to equivalence point after boiling? Assume no change in volume during boiling. 1 4 . H PO is a tri basic acid and one of its salt is NaH PO . What volume of 1 M NaOH solution should be 34 24 added to 12 g of NaH PO to convert it into Na PO ? 24 34 1 5 . 1.64 g of mixture of CaCO3 and MgCO3 was dissolved in 50 mL of 0.8 M HCl. The excess of acid required 16 mL of 0.25 M NaOH for neutralization. Calculate the percentage of CaCO and MgCO in the sample. 33 1 6 . 1.5 g of chalk were treated with 10 mL of 4N – HCl. The chalk was dissolved and the solution made to 100 mL, 25 mL of this solution required 18.75 mL of 0.2 N – NaOH solution for complete neutrali- sation. Calculate the percentage of pure CaCO in the sample of chalk? 3 1 7 . A solution contains Na CO and NaHCO . 20 mL of this solution required 4 mL of 1N – HCl for titration 23 3 with Ph indicator. The titration was repeated with the same volume of the solution but with MeOH. 10.5 mL of 1 – N HCl was required this time. Calculate the amount of Na CO & NaHCO . 23 3 1 8 . A solution contains a mix of Na CO and NaOH. Using Ph as indicator 25 mL of mixture required 19.5 23 mL of 0.995 N HCl for the end point. With MeOH, 25 mL of the solution required 25 mL of the same HCl for the end point. Calculate g/L of each substance in the mixture. N 1 9 . 200 mL of a solution of mixture of NaOH and Na CO was first titrated with Ph and HCl. 17.5 mL 23 10 of HCl was required for end point. After this MeOH was added and 2.5 mL of same HCl was again required for next end point. Find out amounts of NaOH and Na CO in the mix. 23 2 0 . A solution contains Na CO and NaHCO . 10 mL of this requires 2 mL of 0.1 M H SO for neutralisation 23 3 24 using Ph indicator. MeOH is then added when a further 2.5 mL of 0.2 M H SO was needed. Calculate 24 strength of Na CO and NaHCO . 23 3 2 1 . A sample containing Na CO & NaOH is dissolved in 100 mL solution. 10 mL of this solution requires 23 25 mL of 0.1 N HCl when Ph is used as indicator. If MeOH is used as indicator 10 mL of same solution requires 30 mL of same HCl. Calculate % of Na CO and NaOH in the sample. 23 2 2 . It required 40.05 mL of 1 M Ce4+ to titrate 20 mL of 1 M Sn2+ to Sn4+. What is the oxidation state of the cerium in the product. 2 3 . A volume of 12.53 mL of 0.05093 M SeO reacted with exactly 25.52 mL of 0.1 M CrSO . In the reaction, 24 Cr2+ was oxidized to Cr3+. To what oxidation state was selenium conver ted by the reaction. 2 4 . Pottasium acid oxalate K C O .3H C O .4H O can be oxidized by MnO in acid medium. Calculate the 22 4 22 4 2 4 volume of 0.1 M KMnO reacting in acid solution with one gram of the acid oxalate. 4 2 5 . A 1.0 g sample of H O solution containing x% H O by mass requires x cm3 of a KMnO solution for 22 22 4 complete oxidation under acidic conditions. Calculate the normality of KMnO solution. 4
2 6 . Metallic tin in the presence of HCl is oxidized by K Cr O to stannic chloride, SnCl . What volume of 2 27 4 deci-normal dichromate solution would be reduce by 1 g of tin. 2 7 . 5 g sample of brass was dissolved in one litre dil. H SO . 20 mL of this solution were mixed with KI, liberating 24 I2 and Cu+ and the I2 required 20 mL of 0.0327 N hypo solution for complete titration. Calculate the percentage of Cu in the alloy. 2 8 . 0.84 g iron ore containing x percent of iron was taken in a solution containing all the iron in ferrous condition. The solution required x mL of a dichromatic solution for oxidizing the iron content to ferric state. Calculate the strength of dichromatic solution. 2 9 . The neutralization of a solution of 1.2 g of a substance containing a mixture of H C O .2H O, KHC O .H O 22 4 2 24 2 and different impurities of a neutral salt consumed 18.9 mL of 0.5 N NaOH solution. On titration with KMnO solution, 0.4 g, of the same substance needed 21.55 mL of 0.25 N KMnO . Calculate the % 44 composition of the substance. 3 0 . 50 g of a sample of Ca(OH) is dissolved in 50 mL of 0.5 N HCl solution. The excess of HCl was titrated 2 with 0.3 N – NaOH. The volume of NaOH used was 20cc. Calculate % purity of Ca(OH)2. 3 1 . One g of impure sodium carbonate is dissolved in water and the solution is made up to 250 mL. To 50 mL of this made up solution, 50 mL of 0.1 N – HCl is added and the mix after shaking well required 10 mL of 0.16 N – NaOH solution for complete titration. Calculate the % purity of the sample. 3 2 . What amount of substance containing 60% NaCl, 37% KCl should be weighed out for analysis so that after the action of 25 mL of 0.1 N AgNO solution, excess of Ag+ is back titrates with 5 mL of NH SCN 34 solution. Given that 1 mL of NH SCN = 1.1. mL of AgNO . 43 3 3 . A bottle labelled with \"12 V H O \" contain 700 mL solution. If a student mix 300 mL water in it what 22 is the g/litre strength & normality and volume strength of final solution. 3 4 . 50 mL of an aqueous solution of H O were treated with an excess of KI solution and dilute H SO , the 22 24 liberated iodine required 20 mL of 0.1 N Na2S2O3 solution for complete interaction. Calculate the con- centration of H O in g/. 22 3 5 . 100 kg hard water contains 5 g MgSO . Find hardness. 4 3 6 . One litre hard water contains 1 mg CaCl and 1 mg MgSO . Find hardness. 24 3 7 . Calculate the hardness of water sample which contains 0.001 mol MgSO per litre of water. 4 3 8 . A solution of a 0.4 g sample of H O reacted with 0.632 g of KMnO in the presence of sulphuric acid. 22 4 Calculate the percentage purity of the sample of H O . 22 3 9 . 5 litre of a solution of H O with x N strength is diluted to 5.5 litre. This 5.5 litre H O solution gives 22 22 28 litre O at NTP. Find the value of x. 2 4 0 . Calculate the amount of lime Ca(OH) required to remove the hardness in 60 litre of pond water containing 2 1.62 mg of calcium bicarbonate per 100 mL of water. 4 1 . 10 g sample of bleaching powder was dissolved into water to make the solution one litre. To this solution 35 mL of 1.0 M Mohr salt solution was added containing enough H2SO4. After the reaction was complete, the excess Mohr salt required 30 mL of 0.1 M KMnO for oxidation. Find out the % of available Cl 42 approximately is (mol wt. 71).
4 2 . Calculate the amount (in milligrams) of SeO –2 in solution on the basis of following data 20 mL of M/60 3 solution of KBrO was added to a definite volume of SeO –2 solution. The bromine evolved was removed 33 by boiling and excess of KBrO was back titrated with 5 mL of M/25 solution of NaAsO . The reactions 32 are given below. (Atomic mass of K = 39, Br = 80, As = 75, Na = 23, O = 16, Se = 79) (a) SeO –2 + BrO – + H+ S e O –2 + Br + HO 3 3 4 2 2 (b) BrO3– + AsO2– + H2O Br– + AsO4–3 + H+ 4 3 . A 1.0 g sample of Fe O solid of 55.2% purity is dissolved in acid and reduced by heating the solution 23 with zinc dust. The resultant solution is cooled and made upto 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of moles of electrons taken up by the oxidant in the reaction of the above titration. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1. (a) +3 (b) +5 (c) +6 (d) +2 (e) 8/3 or (2 and 3) (f) +3 (g) +2 (h) +2 (i) 200/93 = 2.15 2. (a) S O 2–(aq) + 6Al (s) + 20 H+ 4 HS (aq) + 6Al3+ (aq) + 6H O 46 2 2 (b) 6S O 2–(aq) + Cr O 2– (aq) + 14 H+ 3 S O 2– (aq) + 2 Cr3+ (aq) + 7H O 23 27 46 2 (c) 14ClO –(aq) + 3As S (s) + 18 HO 14 Cl– (aq) + 6H AsO – (aq) + 9HSO – (aq) + 15H+ 3 23 2 24 4 (d) 7IO –(aq) + 6Re (s) + 3H O 6 ReO – (aq) + 7I – (aq) + 6H+ 3 2 4 (e) 30HSO –(aq) + As (s) + 10 Pb O (s) + 26H+ 30 PbSO (s) + 4H AsO – (aq) + 24H O 4 4 34 4 24 2 (f) 3HNO (aq) HNO + 2NO (g) + HO 2 3 2 3. (a) 3C H O 2–(aq) + 5ClO – (aq) + 18 OH– 12 CO 2– (aq) + 5 Cl– (aq) + 15H O 446 3 3 2 (b) 11Al (s) + 3BiONO (s) + 21H O + 11OH– 3Bi (s) + 3NH (aq) + 11Al (OH) –(aq) 3 2 3 4 (c) 4H O (aq) + Cl O (aq) + 2OH– 2ClO – (aq) + 4O (g) + 5H O 22 27 2 2 2 (d) Tl O (s) + 4NH OH (aq) 2TlOH (s) + 2N (g) + 5H O 23 2 22 (e) Cu(NH ) 2+ (aq) + S O 2– (aq) + 4OH– 2SO 2– (aq) + Cu (s) + 4NH (aq) + 2H O 34 24 3 3 2 (f) 3Mn(OH) (s) + 2MnO – (aq) 5MnO (s) + 2H O + 2OH– 2 4 2 2 4. 2, 41 5. (i) 4.0, (ii) 0.0040, (iii) 0.224, (iv) 56.00%, (v) 0.02M, (vi) 0.0008 mol 6. M1 , M2 , M3 7. 1 : 7 : 4 8. 0.06 N and 0.03 M 9 . 537.6 mL 685 10. KOH = 35%, Ca(OH) = 65% 11. V = 157.89 mL 1 2 . 20.72 % 2 13. V = 25 mL 14. 200 mL 15. MgCO = 51.22%, CaCO = 48.78 % 33 16. 83.33 17. 0.424 g ; 0.21g 18. 23.2 g, 22.28 g 1 9 . 0.06 g ; 0.0265 g 20. 4.24 g/L ; 5.04 g/L 21. 39.85% ; 60.15% 22. +3 23. zero 24. V = 31.68 mL 25. 0.588 N 2 6 . 337 mL 27. 41.53 % 28. 0.15 N 29. H C O .2H O = 14.36%, KH O .H O=81.71% 30. 1.406 % 31. 90.1% 22 4 2 24 2 32. 0.1281 g 33. 25.5 g/L, 1.5 N, 8.4 V 34. 0.68 g/L 3 5 . 41.66 ppm 36. 1.734 ppm 37. 100 ppm 38. 85% 39. x = 1 40. 0.444 g 41. 7.1% 42. 84mg 43. 6
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . 1.2475 g of crystalline copper sulphate was dissolved in water and excess of KI was added. The liberated iodine consumed 50 mL N/10 Na2S2O3 solution to reach the end point of the titration. Calculate the number of water molecules of hydration in crystalline copper sulphate salt. 2 . A 1g sample of K2Cr2O7 containing some inert material was entirely reduced with conc. HCl. The chlorine liberated was passed through hot solution of NaOH at 800C, and it completely diproportionates to form ClO3 and Cl. This NaClO3 was isolated and its reduction with KI (aq) liberated iodine, giving Cl.The iodine thus liberated required 100 mL of decinormal hypo solution for complete titration. What is the percentage purity of the dichromate sample? 3 . 2.5g of mixture of crystalline oxalic acid (H2C2O4. 2H2O) and sodium oxalate (Na2C2O4) was dissolved in 100 mL of water. 50 mL of this solution was titrated against N/10 NaOH solution when 119.05 mL of the base was found necessary to reach the end point with phenolphthalein as the indicator. 1g of the mixture was dissolved in water and the solution titrated against N/10 KMnO4 in the presence of dil. H2SO4. What is the volume of KMnO4 needed for getting the end point with 0.5g of the mixture? 4 . 25 mL of a solution containing HCl was treated with excess of M/5 KIO3 and KI solution of unknown concentration where I2 liberated is titrated against a standard solution of 0.021 M Na2S2O3 solution whose 24 mL were used up. Find the strength of HCl and volume of KIO3 solution consumed : 5 . 0.6213 g of sample contains an unknown amount of As2O3. The sample was treated with HCl resulting information of AsCl3 (g) which was distilled into a beaker of water. The hydrolysis reaction is as follows : AsCl3 + 2H2O HAsO2 + 3H+ + 3Cl– The amount of HAsO2 was determined by titration with 0.04134 M I2, requiring 23.04 mL to reach the equivalence point. The redox products in the titration were H3AsO4 and I–. Find the amount of KMnO4 needed to oxidize As in As2O3 to its maximum possible oxidation state in acidic medium. 6 . A sample of steel weighing 0.6 g and containing S as an impurity was burnt in a stream of O2, when S was converted to its oxide SO2.SO2 was then oxidized to SO – – by using H2O2 solution containing 4 30 mL of 0.04 M NaOH. 22.48 mL of 0.024 M HCl was required to neutralize the base remaining after oxidation. Calculate the % of S in the sample : 7 . In the presence of fluoride ion, Mn2+ can be titrated with MnO4–, both reactants being converted to a complex of Mn(III). A 0.545 g sample containing Mn3O4 was dissolved and all manganese was converted to Mn2+. Titration in the presence of fluoride ion consumed 31.1 mL of KMnO4 that was 0.177 N against oxalate. (a) write a balanced chemical equation for the reaction, assuming that the complex is MnF –. 4 (b) what was the % of Mn3O4 in the sample ? 8 . A mixture of two gases, H2S and SO2 is passed through three beakers successively. The first beaker contains Pb2+ ions, which absorbs S2– forming PbS. The second beaker contains 25 mL of 0.0396 N I to oxidize 2 SO2 to SO42–. The third contains 10 mL of 0.0345 N thiosulphate solution to retain any I2 carried over from the second absorber. A 25 L gas sample was passed through the apparatus followed by an additional amount of N to sweep last traces of SO from first and second absorber. The solution from the first absorber 22 was made acidic and treated with 20 mL of 0.0066 M K2Cr2O7 which converted S2– to SO2. The excess dichromate was reacted with solid KI and the liberated iodine required 7.45 mL of 0.0345 N Na2S2O3 solution. The solutions in the second and third absorbers were combined and the resultant iodine was titrated with 2.44 mL of the same thiosulphate solution. Calculate the concentrations of SO2 and H2S in mg/L of the sample :
9 . 1 g of a moist sample of a mixture of KClO3 and KCl was dissolved in water and made upto 250 mL. 25 mL of this solution was treated with SO2 to reduced chlorate into chloride and the excess SO2 was boiled off. When the total chloride was precipitated, 0.1435 g of AgCl was obtained. In another experiment 25 mL of the original solution was treated with 30 mL of 0.2 N solution of FeSO4 and unreacted FeSO4 required 37.5 mL of 0.08 N solution of an oxidizing agent for complete oxidation. Calculate the molar ratio of chlorate and chloride in the given mixture. Fe2+ reacts with ClO3– according to equation : C lO – + 6Fe2+ + 6H+ Cl– + 6Fe3+ + 3H2O 3 Also calculate the mass percent of moisture present in the moist sample. 1 0 . A steel sample is to be analysed for Cr and Mn simultaneously. By suitable treatment the Cr is oxidized to Cr2O 2– and the Mn to MnO4–. A 10.00 g sample of steel is used to produce 250.0 mL of a solution 7 containing Cr2O72– and MnO4–. A 10.00 mL portion of this solution is added to a BaCl2 solution and by proper adjustment of the acidity, the chromium is completely precipitated as BaCrO4 ; 0.0549 g is obtained. A second 10.00 mL portion of this solution requires exactly 15.95 mL of 0.0750 M standard Fe2+ solution for its titration (in acid solution). Calculate the % of Mn and % of Cr in the steel sample. 1 1 . 1.16 g CH3(CH2)nCOOH was burnt in excess air and the resultant gases (CO2 and H2O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part required 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Find the value of n and the amount of excess NaOH solution taken initially. 1 2 . A 1.5 g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume made up to 250 mL. A 20 mL portion of this solution was then mixed with 30 mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of CaCl2 just needed for precipitation of oxalate as CaC2O4. Solution was filtered off and filtrate was finally titrated against 0.1 M HCl solution. 8.0 mL of acid was required to reach the equivalence point. At last, the above neutral solution was treated with excess of AgNO3 solution and AgCl obtained was washed, dried and weighed to be 0.4305 g. De- termine mass percentage of oxalic acid in the original sample : 1 3 . A 1 g sample containing NaOH as the only basic substance and some inert impurity was left exposed to atmosphere for a very long time so that part of NaOH got converted into Na2CO3 by absorbing CO2 from atmosphere. The resulting sample was dissolved in water and volume made upto 100 mL. A 20 mL portion of this solution required 16 mL 0.25 M HCl solution to reach the equivalence point when methyl orange was used as indicator. In a separate analysis, 20 mL portion of the same solution was taken along with phenolphthalein indicator and mixed with 50 mL of 0.1 M HCl solution. An additional 9.00 mL 0.1 M Ba(OH)2 solution was required to just restore the pink colour of solution. Determine mass percentage of NaOH in the original sample and mass percentage of Na2CO3 in the sample after exposure to atmosphere. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1. 5 2 . 58.8% 3 . 77.45 mL 4. VKIO3 0.42 m L , [HCl] = 0.0168 N 5 . 0.06 g 6 . 1.76 % 7. 40.77% 9. ClO3–/Cl– = 1, 1.5% moisture by mass 8 . 0.12 mg H S/L, 0.718 mg SO /L 11. n = 4, NaOH = 6.4 g 22 1 0 . Cr = 2.821% , Mn = 1.498% 12 . 82.5 1 3 . 80, 36.05 %
EXERCISE–05 [A] PREVIOUS YEARS QUESTIONS 1 . MnO4– is good oxidising agent in different medium changing to - [AIEEE-02] (4) 2, 6, 4, 3 M n O – Mn2+ 4 Mn O 2– 4 MnO2 Mn2O3 Changes in oxidation number respectively are - (1) 1, 3, 4, 5 (2) 5, 4, 3, 2 (3) 5, 1, 3, 4 2 . Oxidation number of Cl in CaOCl2 (bleaching powder is ) [AIEEE-02] (1) Zero, since it contains Cl2 (2) –1, since it contains Cl– (3) +1, since it contains ClO– (4) +1 and –1 since it contains ClO– and Cl– 3 . Which of the following is a redox [AIEEE-02] (1) 2NaAg(CN)2 + Zn Na2Zn (CN)4 + 2 Ag (2) BaO2 + H2SO4 BaSO4 + H2O2 (3) N2O5 + H2O 2HNO3 (4) AgNO3 + KI AgI + KNO3 4 . In the coordination compound, K4[Ni (CN)6], the oxidation state of nickel is [AIEEE-03] (1) +1 (2) +2 (3) –1 (4) 0 5 . The oxidation state of Cr in [Cr(NH3)4Cl2]+ is - [AIEEE-05] (1) +2 (2) +3 (3) 0 (4) +1 6 . The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is - [AIEEE-05] (1) +6 (2) +4 (3) +3 (4) +2 7. Which of the following chemical reaction depicts the oxidizing behaviour of H2SO4 ? [AIEEE-06] (1) Ca(OH)2 + H2SO4 CaSO4 + 2H2O (2) NaCl + H2SO4 NaHSO4 + HCl (3) 2PCl5 + H2SO4 2POCl3 + 2HCl + SO2Cl2 (4) 2HI + H2SO4 I2 + SO2 + 2H2O PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5[A] Que. 1 2 3 4 5 6 7 Ans 3 4 1 2 2 3 4
EXERCISE–05 [B] PREVIOUS YEARS QUESTIONS 1 . The oxidation number of phosphorus in Ba(H PO ) is : [JEE 1990] 2 22 (A) +3 (B) +2 (C) +1 (D) –1 2 . The number of electrons to balance the following equation :- [JEE 1991] NO3– + 4H+ + e– 2H2O + NO is (A) 5 (B) 4 (C) 3 (D) 2 3 . What is the volume strength of 1.5 N H O : [JEE 1991] 22 (A) 4.8 (B) 8.4 (C) 3.0 (D) 8.0 4 . The oxidation states of the most electronegative element in the products of the reaction of BaO with 2 dilute H SO . [JEE 1991] 24 (A) 0 and –1 (B) –1 and –2 (C) –2 and 0 (D) –2 and +2 5 . For the redox reaction, [JEE 1992] MnO – + C O 2– + H+ Mn2+ + CO + HO 4 24 2 2 the correct coefficients of the reactants for the balanced reaction are : M n O – C O 2 – H+ 4 4 2 (A) 2 5 16 (B) 16 5 2 (C) 5 16 2 (D) 2 16 5 6 . The number of mole of KMnO that will need to react completely with one mole ferrous oxalate in acidic 4 solution is : [JEE 1997] (A) 2/5 (B) 3/5 (C) 4/5 (D) 1 7 . The number of mole of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is : [JEE 1997] (A) 2/5 (B) 3/5 (C) 4/5 (D) 1 8. The equivalent mass of MnSO is half its molecular mass when it is converted to : [JEE 1998] 4 (A) Mn2O3 (B) MnO2 (C) MnO4– (D) MnO 2– 4 9 . The oxidation number of sulphur in S , S F and H S respectively are : [JEE 1999] 8 22 2 (A) 0, +1 and –2 (B) +2, +1 and –2 (C) 0, +1 and +2 (D) –2, +1 and –2 1 0 . The normality of 0.3 M phosphorus acid (H PO ) is : [JEE 1999] 33 (A) 0.1 (B) 0.9 (C) 0.3 (D) 0.6 1 1 . Among the following species in which oxidation state of the element is +6 : [JEE 2000] (A) MnO4– (B) Cr(CN)63– (C) NiF62– (D) CrO2Cl2 [JEE 2001] 1 2 . Oxidation number of iron in Na2 [Fe(CN)5 NO] is : (A) +2 (B) +3 (C) +8/3 (D) none of these 1 3 . An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is : [JEE 2001] (A) 40 mL (B) 20 mL (C) 10 mL (D) 4 mL 1 4 . How many moles of electron weigh one kilogram : [JEE 2002] (A) 6.023 × 1023 (B) 1 1031 (C) 6.023 1054 (D) 1 108 9.108 9.108 9.108 6.023 [JEE 2003] 1 5 . Which has maximum number of atoms : (A) 24 g of C (12) (B) 56 g of Fe (56) (C) 27 g of Al (27) (D) 108 g of Ag (108) 1 6 . In basic medium I– oxidises by MnO – . In this process I – replaces by : [JEE 2004] 4 (A) IO3– (B) I2 (C) IO4– (D) IO–
1 7 . Amongst the following, the pair having both the metals in their highest oxidation state is :[JEE 2004] (A) [Fe(CN)6]3– and [Co(CN)6]3– (B) [CrO2Cl2] and [MnO4– ] (C) TiO2 and MnO2 (D) [MnCl4]2– and [NiF6] –2 1 8 . O does not oxidise : [JEE 2005] 3 (A) KI (B) FeSO4 (C) KMnO4 (D) K2MnO4 1 9 . A 5.0 cm3 solution of H O liberates 0.508 g of iodine from an acidified KI solution. Calculate the strength 22 of H O solution in terms of volume strength at STP. [JEE' 1995] 22 2 0 . A 3.00 g sample containing Fe O , Fe O and an inert impure substance, is treated with excess of KI solution 34 23 in presence of dilute H2SO4. The entire iron is converted into Fe2+ along with the liberation of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the diluted solution require 11 mL of 0.5 M Na S O 22 3 solution to reduce the iodine present. A 50 mL of diluted solution after complete extraction of the iodine requires 12.80 mL of 0.25 KMnO solution in dilute H SO medium for the oxidation of Fe2+. Calculate 4 24 the percentages of Fe O and Fe O in the original sample. [JEE 2000] 23 34 2 1 . One litre of a mixture of O and O at NTP was allowed to react with an excess of acidified solution of 23 KI. The Iodine liberated required 40 mL of M/10 sodium thiosulphate solution for titration. What is the percent of ozone in the mixture ? Ultraviolet radiation of wavelength 300 nm can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? [JEE 97,5] 2 2 . A sample of hard water contains 96 ppm of SO24– and 183 ppm of HCO–3 , with Ca2+ as the only cation. How many moles of CaO will be required to remove HCO– from 1000 kg of this water? If 1000 kg of 3 this water is treated with the amount of CaO calculate above, what will be the concentration (in ppm) of residual Ca2+ ions (Assume CaCO to be completely insoluble in water)? If the Ca2+ ions in one litre of 3 the treated water are completely exchanged with hydrogen ions, what will be its pH (one ppm means one part of the substance in one million part of water, weight / weights)? [JEE' 1997] 2 3 . An aqueous solution containing 0.10 g KIO (formula wt. 214.0) was treated with an excess of KI solution. 3 The solution was acidified with HCl. The liberated I consumed 45.0 mL of thiosulphate solution to decolourise 2 the blue starch – iodine complex. Calculate the molarity of the sodium thiosulphate solution.[JEE 1998] 2 4 . How many millilitre of 0.5 M H SO are needed to dissolve 0.5 g of copper II carbonate ?[JEE 1999] 24 2 5 . Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO4 (20 mL) acidified with dilute H SO . The same volume of KMnO solution is just decolorized by 10 mL of MnSO in neutral medium 24 4 4 simultaneously forming a dark brown precipitate of hydrated MnO . The brown precipitate is dissolved 2 in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H SO . Write the 24 balanced equations involved in the reactions and calculate the molarity of H O . [JEE 2001] 22 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5[B] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C C B B A B A B A D D A A D A Que. 16 17 18 Ans. A B C 1 9 . 4.48 2 0 . Fe O = 49.33 %, Fe O = 34.8% 2 1 . 6.57% O3 (by weight), 1.2 × 1021 photons 23 34 2 3 . 0.0623 M 2 2 . 1.5, 40 ppm, pH = 2.6989 2 4 . 8.097 mL 2 5 . 0.1 M
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Alkaline earth metals (group 2 or IIA elements) differ from group 12 (or IIB) elements in the electronic configuration of their : (A) Antipenultimate shell (B) Innermost shell (C) Outermost shell (D) Penultimate shell 2 . The first ionization enthalpy of magnesium is lower than the first ionization enthalpy of : (A) Lithium (B) Sodium (C) Calcium (D) Beryllium 3 . Chemical A is used for water softening to remove temporary hardness. A reacts with Na2CO3 to generate caustic soda. When CO2 is bubbled through A, it turns cloudly. What is the chemical formula of A : (A) CaCO3 (B) CaO (C) Ca(OH)2 (D) Ca(HCO3)2 4 . The substance not likely to contain CaCO3 is : (A) Calcined gypsum (B) Sea shells (C) Dolomite (D) A marble statue 5 . A metal M readily forms water soluble sulphate MSO4, water insoluble hydroxide M(OH)2 and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. The M is : (A) Be (B) Mg (C) Ca (D) Sr 6 . A chloride dissolves appreciably in cold water. When placed on a Pt wire in Bunsen flame, no distinctive colour is noted. Which cation could be present? (A) Be2+ (B) Ba2+ (C) Pb2+ (D) Ca2+ 7 . The hydroxide which is best soluble in water is : (A) Ba(OH)2 (B) Mg(OH)2 (C) Sr(OH)2 (D) Ca(OH)2 8 . What is X in the following reaction? MgCl2 + 2 H2O X + 2 HCl + H2O (A) MgO (B) Mg (C) Mg(OH)2 (D) Mg(OH) Cl 9 . (Yellow ppt) T K2CrO4 X dil. HCl Y (Yellow ppt) + Z (pungent smelling gas) If X gives green flame test. Then, X is : (A) MgSO4 (B) BaS2O3 (C) CuSO4 (D) PbS2O3 1 0 . The correct statement is/are : (A) BeCl2 is a covalent compound (B) BeCl2 is an electron deficient molecule (C) BeCl2 can form dimer (D) The hybrid state of Be in BeCl2 is sp2 1 1 . The reaction of an element A with water produces combustible gas B and an aqueous solution of C. When another substance D reacts with this solution C also produces the same gas B. D also produces the same gas even on reaction with dilute H2SO4 at room temperature. Element A imparts golden yellow colour to Bunsen flame. Then A, B, C and D may be identified as : (A) Na, H2 NaOH and Zn (B) K, H2, KOH and Zn (C) K, H2, NaOH and Zn (D) Ca, H2, CaCOH2 and Zn 1 2 . An alkaline earth metal (M) gives a salt with chlorine, which is insoluble in water at room temperature but soluble in boiling water. It also forms an insoluble sulphate whose mixture with a sulphide of a transition metal is called 'lithopone'a white pigment. Metal M is : (A) Ca (B) Mg (C) Ba (D) Sr 1 3 . In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam: (A) Hg is more inert than Pt (B) More voltage is required to reduce H+ at Hg than at Pt (C) Na is dissolved in Hg while it does not dissolve in Pt (D) Conc. of H+ ions is larger when Pt electrode is taken
1 4 . The correct sequence of increasing covalent character is represented by – (A) BeCl2 < NaCl < LiCl (B) NaCl < LiCl < BeCl2 (C) BeCl2 < LiCl < NaCl (D) LiCl < NaCl < BeCl2 1 5 . The paramagnetic species is : (A) KO2 (B) SiO2 (C) TiO2 (D) BaO2 1 6 . The pair of amphoteric hydroxides is :- (A) Al(OH) , LiOH (B) Be(OH) , Mg(OH) 3 22 (C) B(OH) , Be(OH) (D) Be(OH) , Zn(OH) 32 22 1 7 . Maximum thermal stability is shown by (A) MgCO (B) CaCO (C) SrCO (D) BaCO 3 3 3 3 1 8 . Stable oxide is obtained by heating the carbonate of the element (A) Li (B) K (C) Na (D) Rb 1 9 . The stable superoxide is formed by the element (A) Li (B) Na (C) K (D) Ca 2 0 . The metallic lustre exhibited by sodium is explained by (A) diffusion of sodium ions (B) oscillation of loose electrons (C) excitation of free protons (D) existence of body centred cubic lattice 2 1 . A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively (A) H2 , O2 (B) O2 , H2 (C) O2 Na (D) O2, SO2 2 2 . The hydration energy of Mg2+ is greater than that of (A) Al3+ (B) Na+ (C) Be2+ (D) Mg3+ 2 3 . Calcium is obtained by the (A) electrolysis of molten calcium chloride (B) electrolysis of a solution of CaCl in water 2 (C) reduction of CaCl with carbon 2 (D) roasting of limestone 2 4 . The material used in photoelectric cells contains – (A) Cs (B) Si (C) Sn (D) Ti 2 5 . Four alkali metals A, B, C and D are having respectively standard reduction potentials as –3.05, –1.66, –0.40 and 0.80 V. Which one will be the most reducing agent ? (A) A (B) B (C) C (D) D 2 6 . Which of the following imparts violet colouration to the Bunsen burner non-luminous flame (A) NaCl (B) BaCl (C) CaCl (D) KCl 2 2 2 7 . Which one of the following is most basic ? (A) Al O (B) MgO (C) SiO (D) P O 23 2 25 2 8 . Molten sodium is used in nuclear reactors to (A) absorb neutrons in order to control the chain reaction (B) slow down the fast neutrons (C) absorb the heat generated by nuclear fission (D) extract radio-isotopes produced in the reactor
2 9 . Bone ash contains (A) CaO (B) CaSO 4 (C) Ca (PO ) 3 42 (D) Ca(H PO ) 2 42 3 0 . Which of the following does not illustrate the anomalous properties of Li ? (A) The m.p. and b.p. of Li are comparatively high (B) Li is much softer than the other I group metals (C) Li forms a nitride Li N unlike group I metals 3 (D) The ion of Li and its compounds are more heavily hydrated than those of the rest of the group 3 1 . Of the following the commonly used as a laboratory desicator is (A) Na CO 23 (B) CaCl 2 (C) NaCl (D) None of the above 3 2 . The increasing order of solubility is (A) CaCO .KHCO , NaHCO 33 3 (B) NaHCO , KHCO , CaCO 3 33 (C) KHCO3, NaHCO3, CaCO3 (D) CaCO , NaHCO , KHCO 3 33 3 3 . Which one of the following compounds gives methane on treatment with water ? (A) Al4C3 (B) CaC2 (C) VC (D) SiC 3 4 . Sodium loses its lustre on exposure to air due to formation of – (A) Na O, NaOH and Na CO 2 23 (B) Na O and NaOH 2 (C) Na O and Na CO 2 23 (D) NaOH and Na CO 23 3 5 . Which of the following hydride is covalent and polymeric :- (A) CaH 2 (B) BeH 2 (C) NaH (D) BaH 2 CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. D D C A A A A A B A,B,C A C D B A Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. D D A C B A BAAADB C CB 34 35 Que. 31 32 33 Ans. B D A A B
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT NSWERS) 1 . Which of the following is incorrect? (A) Mg burns in air releasing dazzling light rich in UV rays. (B) CaCl2 · 6 H2O when mixed with ice gives freezing mixture. (C) Mg cannot form complexes (D) Be can form complexes due to its very small size. 2 . On dissolving moderate amount of sodium metal in liquid NH at low temperature, which one of the following 3 does not occur (A) Blue coloured solution is obtained. (B) Na+ ions are formed in the solution. (C) Liquid NH becomes good conductor of electricity. 3 (D) Liquid ammonia remains diamagnetic. 3 . The minimum equivalent conductance in fused state is shown by – (A) MgCl (B) BeCl (C) CaCl (D) SrCl 2 2 2 2 4 . The metal which cannot be produced on reduction of its oxide by aluminium is (A) K (B) Mn (C) Cr (D) Fe 5 . Magnesium on reaction with very dilute HNO gives 3 (A) NO (B) N O (C) H (D) NO 2 2 2 6 . The alkali metal that reacts with nitrogen directly to form nitride is (A) Li (B) Na (C) K (D) Rb 7 . Which of the following statement is/are false for alkali metals ? (A) Lithium is the strongest reducing agent (B) Na is amphoteric in nautre (C) Li+ is exceptionally small (D) All alkali metals give blue solution in liquid ammonia 8 . Amongst LiCl, RbCl, BeCl and MgCl , the compounds with the greatest and least ionic character respectively 22 are :- (A) LiCl, RbCl (B) RbCl, BeCl2 (C) RbCl, MgCl2 (D) MgCl2, BeCl2 (C) thiocarbonate (D) thiocyanate 9 . K CS can be called potassium 23 (A) sulphocyanide (B) thiocarbide 1 0 . Anhydrous MgCl2 can be prepared by heating MgCl2.6H2O (A) in a current of dry HCl gas (B) with carbon (C) until it fuses (D) with lime 1 1 . Oxygen ions structure in its peroxide, superoxide, ozonide : (A) O —, O 2, O —2 (B) O —2, O -—, O — (C) O —2, O—2, O — (D) O —, O —3, O —2 2 23 2 23 23 22 3
1 2 . In presence of iron, alkali metal react with liquid ammonia and form (A) Metal mixture + H2 (B) Iron metal mixture + H2 (C) Metal mixture (D) Metal amide + H2 1 3 . The ionic conductance of following cation in a given concentration are in the order (A) Li+ < Na+ < K+ < Rb+ (B) Li+ > Na+ > K+ > Rb+ (C) Li+ < Na+ > K+ > Rb+ (D) Li+ = Na+ < K+ < Rb+ 1 4 . Which of the following does not give an oxide on heating – (A) MgCO (B) Li CO (C) ZnCO (D) K CO 3 23 3 23 1 5 . On heating sodium metal in the current of dry ammonia leads to the formation of which gas– (A) NaNH (B) NaN (C) NH (D) H 2 3 3 2 1 6 . On allowing ammonia solution of s-block metals to stand for a long time, blue colour becomes fade. The reason is:- (A) Formation of NH3 gas (B) Formation of metal amide (C) Cluster formation of metal ions (D) Formation of metal nitrate 1 7 . When Na and Li placed in dry air we get :- (A) NaOH, Na O, Li O (B) Na CO , Na O , Li O 22 23 22 2 (C) Na O, Li N, NH 3 (D) Na O, Li O, Li N 23 223 1 8 . The hydride ion H– is stronger base than its hydroxide ion OH–. Which of the following reaction will occur if sodium hydride is dissolved in water:- (A) – + H2O H3O+ – + H2O OH– + H2 H(aq) (B) H(aq) – – (C) H + H2O H2 + O2 (D) H + H2O No reaction 1 9 . Which can not be used to generate H :– 2 (A) Al + NaOH (B) Zn + NaOH (C) Mg + NaOH (D) LiH + H O 2 2 0 . Only those elements of s-block can produce superoxides which have :- (A) High ionisation energy (B) High electronegativity (C) High charge density (D) Low ionisation potential 2 1 . Alum is the name used for all double salts having the composition M I SO .M III(SO ) .24H O. Where MIII 2 42 43 2 stands for Al+3, Cr+3, Fe+3, while MI stands for:- (A) Li+, Cu+, Ag+ (B) Li+, NH +, Na+ (C) Na+, K+, Rb+ (D) Ca+2, Mg+2, Sr+2 4 2 2 . Identify the correct statement - (A) Gypsum contains a lower percentage of Ca than plaster of paris (B) Gypsum is obtained by heating plaster of paris (C) Plaster of paris can be obtained by hydration of gypsum (D) Plaster of paris is obtained by partial oxidation of gypsum 23. In the reaction M + O MO (super oxide) the metal is 2 2 (A) Li (B) Na (C) K (D) Ba
2 4 . Na+ and Ag+ differ in (A) Na CO is thermally stable while Ag CO decomposes into Ag, CO and O 23 23 22 (B) Ag+ forms complexes, Na+ does not (C) NaCl is water soluble, AgCl is insoluble (D) NaBr-yellow and AgBr pale yellow 2 5 . The stability order of oxide, peroxide and superoxide of alkali metal is (A) Normal oxide > super oxide > per oxide (B) Normal oxide > per oxide > super oxide (C) super oxide > per oxide > normal oxide (D) per oxide > normal oxide > super oxide 2 6 . Match list I with list II and choose the correct answer from the codes given below List I List II (A) NaNO (a) Baking soda 3 (b) Chile salt peter (c) Microcosmic salt (B) Na(NH )HPO (d) Washing soda 44 (C) NaHCO 3 (D) Na CO .10H O 23 2 Codes is : A BC D (A) a bc d (B) b ca d (C) c ab d (D) d ab c 2 7 . Which of the following statement is not correct (A) LiOH is amphoteric in nature (B) LiCl is soluble in pyridine (C) Li N is stable while Na N doesn't exist even at room temperature 33 (D) BeO is amphoteric in nature 2 8 . Which of the following statement is correct for s–block elements :- (A) Be has smallest atomic size in II A group (B) Li is most metallic (C) Mg impart red colour to the flame (D) Cs is most reducing in water 2 9 . Which of the following are ionic carbides? (A) CaC2 (B) Al4C3 (C) SiC (D) Be2C 3 0 . Which of the following groups of elements have chemical properties that are most similar : (A) Na, K, Ca (B) Mg, Sr, Ba (C) Be, Al, Ca (D) Be, Ra, Cs
3 1 . Which of the following statements are false? (A) BeCl2 is a linear molecule in the vapour state but it is polymeric in the solid state (B) Calcium hydride is called hydrolith (C) Carbides of both Be and Ca react with water to form acetylene (D) Oxides of both Be and Ca are amphoteric. 3 2 . The incorrect statement(s) is/are : (A) Mg cannot form complexes (B) Be can form complexes due to a very small atomic size (C) The first ionisation potential of Be is higher than that of Mg. (D) Mg forms an alkaline hydroxide while Be forms amphoteric oxides. 3 3 . Na2SO4 is water soluble but BaSO4 is insoluble because : (A) The hydration energy of Na2SO4 is higher than that of its lattice energy (B) The hydration energy of Na2SO4 is less than that of its lattice energy (C) The hydration energy of BaSO4 is less than that of its lattice energy (D) The hydration energy of BaSO4 is higher than that of its lattice energy 3 4 . BeCl2 + LiAlH4 X + LiCl + AlCl3 (A) X is lithium hydride (B) X is BeH2 (C) X is BeCl2 · 2H2O (D) X is LiH 3 5 . X CaCl2 CaCl2 + Y ; the effective ingredient of X is : (A) OCl– (B) Cl– (C) OCl+ (D) OCl2– 3 6 . Which of the following substance(s) is/are used in laboratory for drying purposes? (A) Anhydrous P2O5 (B) Graphite (C) Anhydrous CaCl2 (D) Na3PO4 3 7 . If X and Y are the second ionisation potentials of alkali and alkaline earth metals of same period, then : (A) X > Y (B) X < Y (C) X = Y (D) X << Y 3 8 . X N2 , Y H2O Z (colourless gas) CuSO4 T (blue colour) : Then, substances Y and T are – (A) Y = Mg3N2 and T = CuSO4 · 5H2O (B) Y = Mg3N2 and T = CuSO4 · 4NH3 (C) Y = Mg(NO3)2 and T = CuO (D) Y = MgO and T = CuSO4 · 4NH3 3 9 . When K2O is added to water, the solution becomes basic in nature because it contains a significant concentration of : (A) K+ (B) O2– (C) OH– (D) O 2 2
40. (White ppt) D Na2CO3 A (inKa2cCetricOa4cid B (Yellow ppt) dil. H2SO4 C(White ppt) If A is the metallic salt, then the white ppt. of D must be of (A) Magnesium oxide (B) Red lead (C) Barium carbonate (D) Calcium carbonate 4 1 . Which of the following compounds are paramagnetic in nature? (A) KO2 (B) K2O2 (C) Na2O2 (D) RbO2 4 2 . NaOH(Solid) + CO 200ºC X; product X is : (A) NaHCO3 (B) NaHCO2 (C) HCOONa (D) H2CO3 4 3 . EDTA is used in the estimation of : (A) Mg2+ ions (B) Ca2+ ions (D) Mg2+ ions but not Ca2+ ions (C) Both Ca2+ and Mg2+ ions 44. Na + Al2O3 hightemperature X CwOa2teirn Y; compouind Y is : (A) NaAlO2 (B) NaHCO3 (C) Na2CO3 (D) Na2O2 4 5 . The compound(s) which have –O–O– bond(s) is/are : (A) BaO2 (B) Na2O2 (C) CrO5 (D) Fe2O3 4 6 . KO2 finds use in oxygen cylinders used for space and submarines. The fact(s) related to such use of KO2 is/are : (A) it produces O2 (B) It produces O3 (C) It absorbs CO2 (D) It absorbs both CO and CO2 4 7 . CsBr3 contains : (A) Cs–Br covalent bonds (B) Cs3+ and Br– ions (C) Cs+ and Br3– ions (D) Cs3+ and Br33– ions 4 8 . Fire extinguishers contain : (A) conc. H2SO4 solution (B) H2SO4 and NaHCO3 solutions (C) NaHCO3 solution (D) CaCO3 solution BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C D B A C A BBCABD A DD 19 20 21 22 23 24 25 26 27 28 29 30 Que. 16 17 18 Ans. B D B C D C A C D B B A A A,B,D B,C 34 35 36 37 38 39 40 41 42 43 44 45 Que. 31 32 33 Ans. C,D, A A,C, B A A,C A B B C A,D C C C A,B,C Que. 46 47 48 Ans. A,C C B
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . Magnesium is an essential constituent of chlorophyll, the green colouring matter of plants. 2 . Setting of cement is an endothermic process. 3 . Calcium bicarbonate is known in solid state. 4 . BeH is an ionic hydride. 2 5 . BeCO is thermally stable compounds. 3 6 . In the electrolysis of fused calcium hydride, hydrogen is liberated at cathode. 7 . MgCl .6H O on heating forms MgCl . 22 2 8 . Sodium when heated in excess of oxygen gives sodium oxide. 9 . In group IA of alkali metals, the ionisation potential decreases down the group. Therefore, lithium is a poorer reducing agent in gaseous medium. 1 0 . The softness of group IA metals increases down the group with increasing atomic number. FILL IN THE BLANKS 1 . Anhydrous magnesium chloride is obtained by heating the hydrated salt with................... 2 . Ca(OH) is..................... basic than Mg(OH) . 22 3 . CaH is comercially known as..................... 2 4 . Magnesium burn is air forming ..................... and ..................... 5 . Ba react with cold water ..................... Mg reacts with..................... while Be has..................... with boiling water. 6 . A standard solution of sodium hydroxide cannot be prepared by direct weighing because..................... 7 . Potassium bicarbonate cannot be prepared by solvay process because..................... 8 . Solution of alkali metals in liquid ammonia conducts electricity due to..................... MATCH THE COLUMN 1. Column-I Column-II (A) Hydrolith (p) Contain Ca (B) Nitrolium (q) Used as a fertilizer (C) Dolomite (r) Used to prepare H (D) Pearl's ash 2 (s) Contain potassium 2. Column-I Column-II (A) Solvay process (p) NaCl (q) Na O (B) Evolve CO on heating 2 22 (C) aq. soln. is neutral towards litmus (r) NaHCO 3 (D) Oxone (s) Na CO 23
3. Column-I Column-II (A) Metal sulphate metal oxide + SO + O (p) Ba 2 2 Sr (B) Metal cation + K CrO yellow ppt (q) Na 24 Mg (C) Metal + NH (liquid) blue solution (r) 3 (s) (D) MCl + conc. H SO white ppt. 2 24 ASSERTION & REASON QUESTIONS These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is not a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . S t a t e m e n t - I : Li SO do not form double salt like alum. 24 Because S t a t e me n t - I I : Atomic size of Li is too small. 2 . S t a t e m e n t - I : NaCl when exposed in air it becomes wet. Because S t a t e me n t - I I : NaCl contains hygroscopic impurities like CaCl2, MgCl2 etc. 3 . Statem ent-I : Lithium is the weakest reducing agent among alkali metals. Because S t a t e me n t - I I : In alkali metals I.P. decreases down the group. 4 . S t a t e m e n t - I : BaCO is more soluble in HNO than in plain water. 33 Because Statement-II : Carbonate is a weak base and reacts with the H+ from the strong acid causing the barium salt to dissociate. 5 . S t a t e m e n t - I : BeCl fumes in moist air. 2 Because S t a t e me n t - I I : BeCl reacts with moisture of form HCl gas. 2 COMPREHENSION BASED QUESTIONS Comprehension # 1 A B (oxide) + CO 2 B + HO C 2 C + CO2 A (milky) C + NH Cl D (gas) 4 D + H O + CO E 22 E + NaCl F F Na CO + CO + HO 23 2 2 1 . A is : (A) Ca(HCO ) (B) CaCO 32 3 (C) CaO (D) Na CO 23
2 . B and C are : (A) CaO, Ca(OH) (B) Ca(OH) , CaCO 2 23 (C) CaCO , Ca(OH) (D) Ca(OH) , CaO 32 2 3 . D, E and F are : (A) NH , NH Cl, NH HCO 34 43 (B) NH , NH HCO , NaHCO 343 3 (C) NH HCO , Na CO , NaHCO 4 3 23 3 (D) None Comprehension # 2 Following given passage the five observation regarding alkali metals are mentioned. (i) On exposure to air, sodium hydroxide becomes liquid and after some time it changes to white powder. (ii) In water LiF is least soluble fluorides among fluorides of alkali metal, but its solublity increases as HF is added in aqueous solution. (iii) LiH more stable than NaH when heated separately (iv) When excess of Na S O solution is added to the FeCl solution an intense violet colouration is produced, 22 3 3 but violet colour disappeared shortly (v) Between Na+ and Ag+, Ag+ is stronger Lewis acid. 1 . The explanation of observation (v) is : (A) Because Na+ has inert gas configuration which has greater polarisation power (B) Because Ag+ has inert gas configuration which has greater polarisation power (C) Because Ag+ has pseudo inert gas configuration which has lesser polarisation power (D) Because Ag+ has pseudo inert gas configuration which has greater polarisation power 2 . The explanation for the observation (iv) is : (A) Initialy with FeCl3, Na2S2O3 produce an intense violet colour substance Fe2(S2O)3. But Fe2(S2O3)3 changes to Fe2+ & S4O62– on standing (B) Initially with FeCl3, is reduced to FeCl2 by Na2S2O3. FeCl2 so produced undergo unstable complex formation Fe(S2O3)34– which is violet in colour. (C) Initially with FeCl3, Na2S2O3 produce colloidal Fe which is violet in colour. After sometime, the colloidal suspension changes to the ppt of Fe. (D) There is no reaction 3 . As per observation (iii) LiH is more stable than NaH, because : (A) Due to small size of Li+, the lattice energy of LiH is greater (B) Due to greater size of H–, the lattice energy of LiH is greater (C) LiH is more covalent than NaH (D) Due to greater size of Na+, the lattice energy of NaH is greater. 4 . As per observation (ii) the solubility of LiF increases in the presence of HF, because : (A) The HF further ionises to H+ & F– (B) In the presence of HF, there will be a comon ion effect (C) In the presence HF, F– is converted to HF2 (D) All of the above
5 . The reaction for observation (i) can be explained as – (A) NaOH (S) H2O NaOH(aq) p; H2O N (B) NaOH (S) H2O NaOH(aq) H2O Na2O (S) (C) NaOH (S) H2O NaOH(aq) O2 / air Na2O (S) (D) NaOH (S) H2O NaOH(aq) CO2 Na2CO (S) Comprehension # 3 Na H2O a CO2 B SO2 C Na2S/I2 D Ag+/salt E (complex) 1 . The compound B and C are : (A) Na CO , Na SO 23 24 (B) NaHCO , Na SO 3 24 (C) Na CO , Na SO 23 23 (D) None of these 2 . The compound D is : (A) Na2SO4 (B) Na2S4O6 (C) Na2S2O5 (D) Na2S2O3 3 . Oxidation number of each 'S' atom in compound D : (A) + 2, + 2 (B) + 4, 0 (C) + 6, – 2 (D) + 5, – 1 Comprehension # 4 Alkali metals readily react with oxyacids forming corresponding salts like M CO , MHCO , MNO , M SO 23 3 3 24 etc. with evolution of hydrogen. They also dissolve in liquid NH but without the evolution of hydrogen. The 3 colour of its dilute solution is blue but when it is heated and concentrated then its colour becomes bronze. 1 . Among the nitrate of alkali metals which one can be decomposed to its oxide? (A) NaNO3 (B) KNO3 (C) LiNO3 (D) All of these 2 . Among the carbonates of alkali metals which one has highest stability? (A) Cs2CO3 (B) Rb2CO3 (C) K2CO3 (D) Na2CO3 3 . Which of the following statement about the sulphate of alkali metal is correct? (A) Except Li2SO4 all sulphate of other alkali metals are soluble in water (B) All sulphates of alkali metals except lithium sulphate forms alum. (C) The sulphates of alkali metals cannot be hydrolysed. (D) All of these
4 . Which of the following statement about solution of alkali metals in liquid ammonia is correct? (A) The solution have strong oxidizing properties. (B) Both the dilute solution as well as concentrated solution are paramagnetic in nature (C) Charge transfer is the responsible for the colour of the solution (D) None of these 5 . Which metal bicarbonates does not exist in solid state? (i) LiHCO3 (ii) Ca(HCO3)2 (iii) Zn (HCO3)2 (iv) NaHCO3 (v) AgHCO3 (A) (ii), (iii), (v) (B) (i), (ii), (iii) (C) (i), (ii), (v) (D) (ii), (iii), (iv) MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 1. T 2. F 3. F 4. F 5. F 6. F 7. F 8. F 9. T 10. T Fill in the Blanks 1. Dry HCl 2. More 3. Hydrolyth 4. Mgo, Mg N 5. Vigrowly, slowly, no action 32 6. Absorb moisture & CO from atmosphere 7. Ammonated e– 2 Match the Column 1. (A) p,r ; (B) p,q ; (C) p ; (D) s 2. (A) r,s ; (B) r,s ; (C) p ; (D) q 3. (A) p,q,s ; (B) p,q; (C) p,q,r,s ; (D) p,q Assertion - Reason Questions 1. A 2. A 3. D 4. C 5. A Comprehension Based Questions Comprehension #1 : 1. (B) 2. (A) 3. (B) Comprehension #2 : 1. (D) 2. (A) 3. (A) 4. (C) 5. (D) Comprehension #3 : 1. (C) 2. (D) 3. (C) Comprehension #4 : 1. (C) 2. (A) 3. (D) 4. (D) 5. (A)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Explain the following : (i) The reaction between marble and dilute H SO is not used to prepare carbon dioxide. 24 (ii) Lime water becomes turbid on passing CO though it, but becomes clear when more CO is passed. 22 (iii) Alkaline earth metals have higher melting points than alkali metals. (iv) Beryllium does not exhibit a covalency beyond 4. 2 . PbO is soluble in NaOH and also in HCl. What does it reflect about the nature of PbO ? 22 3 . What happens when : (i) Hot and concentrated caustic soda solution reacts with iodine. (ii) White phosphorus is heated with caustic soda. (iii) Excess of caustic soda reacts with zinc sulphate solution. (iv) Excess of NaOH is added to AlCl solution. 3 4 . Write balanced equation for reaction between (i) Na2O2 and water (ii) KO2 and water (iii) Na2O2 and CO2 5 . Element A bruns in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide . Identify A, B, C and D. 6 . In water LiF is least soluble fluoride among fluorides of alkali metals, but its solubility increases as HF is added in aqueous solution, why? 7 . What happens when CuSO (aq.) is treated with excess of Na S O solution? 4 22 3 8 . Arrange the following in order of increasing .............. (i) Thermal stability BeSO4, MgSO4, CaSO4 (ii) Polarising power Be2+, Mg2+, Ca2+ (iii) Solubility in H2O Be(OH)2, Mg(OH)2, Ca(OH)2 (iv) Covalent nature BeCl2, MgCl2, CaCl2 (v) Hydrolysis nature BeCl2 MgCl2, CaCl2 (vi) Lattice energy CaF2, MgF2, BaF2 (vii) Hydration energy Be2+, Mg2+, Ba2+ (viii) Solubility in water MgF2, BaF2, BeF2 (ix) Basic nature Be, Mg, Ca, Sr 9 . Hydrogen reacts with a metal (A) to give an ionic hydride (B). The metal (A) gives brick red colour with bunsen flame. The hydride formed is commonly known by its trade name. The compound (B) on treating with water gives back H2 and (C). Identify (A), (B) and (C).
CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 . (i) Insoluble CaSO is formed which deposists on the surface of marble and prevents further action of dilute 4 H SO , so the evolution of CO ceases after sometime. 24 2 (ii) Insoluble CaCO is first precipitated which dissolves in excess of CO due to the form of Ca(HCO ) . 3 2 32 Ca(OH) + CO CaCO + H O; 2 2 3 2 (Insoluble) CaCO + HO + CO Ca(HCO ) 3 2 2 32 (Soluble) (iii) Metallic bonding is much stronger in alkaline earth metals as two electrons are present in valence shell. (iv) The outermost energy shell in beryllium is the second. It cannot accommodate more than 8 electrons and hence a covalency limit 4 cannot be exceeded. 2 . PbO are amphoteric nature 2 PbO + 4HCl PbCl + 2H O 2 4 2 PbO + 2NaOH Na PbO + H O 2 23 2 3 . (i) 3I + 6NaOH 5NaI + NaIO + 3H O (ii) P + 3NaOH + 3H O PH +3NaH PO 2 3 2 4 2 3 22 (iii) ZnSO + 2NaOH Zn(OH) +Na SO 4 (iv) AlCl + 3NaOH Al(OH) +3NaCl 4 22 3 3 ppt ppt Zn(OH) +2NaOH Na [ Zn(OH ) 4 Al(OH) + NaOH Na[Al(OH) ] 2 2 3 4 or NaAlO 2 4 . (i) Na O + 2H O 2NaOH + HO 22 2 22 (ii) KO + HO KOH + HO + O 2 2 22 2 (iii) 2Na O + 2CO 2Na CO + O 22 2 23 2 5 . A = Ca, B = Ca N C = Ca(OH) D = NH 32 2 3 6 . In presence of HF, F– is converted into bifluoride ion HF2–, allowing further dissolution of solid LiF. 7. CuSO + Na S O Cu S O 3 + Na SO 4 22 3 24 2 2CuS O + Na S O C u S O + Na S O 23 22 3 24 6 23 Cupric thiosulphate 3Cu2S2O3 + 2Na2S2O3 Na4[Cu6(S2O3)5] Sodium cuprothiosulphate. 8 . (i) BeSO < MgSO < CaSO (ii) Ca2+ < Mg2+ < Be2+ 4 44 (iii) Be (OH) < Mg(OH) < Ca(OH) (iv) CaCl < MgCl < BeCl 22 2 22 (v) CaCl < MgCl < BeCl 2 22 (vi) BaF < CaF < MgF 2 (vii) Ba2+ < Mg2+ < Be2+ 22 (ix) Be < Mg < Ca < Sr (viii) BaF < MgF < BeF 9 . (i) Ca gives brick red colour to flame 2 22 (ii) Ca + H CaH (hydrolith, trade name) 2 2 (A) (B) (iii) CaH + 2H O Ca(OH)2 + 2H 2 2 2 (B) (C)
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . When a gas (A) is passed through dry KOH at low temperature, a deep red coloured compound (B) and a gas (C) are obtained. The gas (A) on reaction with but-2-ene followed by treatment with Zn/H2O yields acetaldehyde. Identify (A), (B) and (C) 2 . A compound (X) imports a golden yellow flame and shows the following reactions: (i) Zinc powder when boiled with a concentrated aqueous solution of (X) dissolves and hydrogen is evolved. (ii) When an aqueous solution of (X) is added to an aqueous solution of stannous chloride, a white precipitate is obtained first which dissolves in excess of solution of (X). Identify (X) and write equations at step (i) and (ii). 3 . A white solid is either Na2O or Na2O2. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid. (i) Identify the substance and explain with balanced equation (ii) Explain what would happen to the red litmus if the white solid were the other compound 4 . (A) is binary compound of a univalent metal. 1.422 g of (A) reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid (B) that formed a hydrated double salt (C) with Al2(SO4). Identify (A),(B) and (C). 5 . Element (M) is a shiny and highly reactive metal (melting point 63ºC) and element (X) is a highly reactive non-metal (melting point – 7.2ºC). They react to form a compound with the empirical formula MX, a colourless, brittle solid that melts at 734ºC. When dissolved in water or when in the molten state, the substance conduct electricity. When chlorine gas is bubbled through an aqueous solution containing (MX), a reddish-brown liquid appears and and Cl– are formed. From these observations, identify M and X. 6 . Name an element which is invariable bivalent and whose oxide is soluble in excess of NaOH and its dipositive ion has a noble gas core. 7 . Out of the elements marked A, B, C, D, E, F, G and H: (a) Which form superoxide? (b) Which form thermally stable carbonate? (c) Which forms strongest base? (d) Which show diagonal relationship? (e) Which forms amphoteric oxide? AB CD EF GH
CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 1 . The gas (A) on treatment with but-2-ene followed by treatment with Zn/H O yields acetaldehyde and thus (A) is 2 ozone (i) O + CH — CH=CH–CH C H — CH—O—CH—CH3 3 3 3 3 OO Mono ozonide 2CH3CHO H2O Zn Acetaldehyde (ii) 5O3 + 2KOH 2KO3 + H2O(g) + 5O2 (A) Potassium ozonide (C) Deep red (B) 2 . (i) Zn + 2NaOH Na2ZnO2 + H2 (X) (ii) (X) is also justified by step 2 reactions: 2NaOH + SnCl2 Sn(OH)2 + 2NaCl (X) White ppt. Sn(OH)2 + 2NaOH Na2SnO2 + 2H2O (Excess) Soluble (X) 3 . (i) Na O + 2H O 2NaOH + HO + [O] 22 2 2 [O] + Litmus White (bleaching) Red (ii) The other compound Na O will give NaOH on dissolution in water. The red litmus will turn to blue. 2 4. 2KO + S K SO Al2(aSqO. 4)3K2SO4 · Al (SO ) · 24H O 2 24 2 43 2 (A) (B) (C) 5 . The given facts suggest M to be potassium (K) and (X) to be bromine (Br ). 2 2K + Br 2KBr 2 (Ionic solid with m. pt. 734°C) 6. Be 7 . (a) E and G (b) C, E and G (c) G (d) A and D (e) B
EXERCISE–05 (A) PREVIOUS YEARS QUESTIONS 1. A metal M readily forms its sulphate MSO4 which is water soluble. It forms oxide MO which becomes inert on heating. It forms insoluble hydroxide which is soluble in NaOH. The metal M is:- [AIEEE-2002] (1) Mg (2) Ba (3) Ca (4) Be 2. KO2 is used in space and submarines because it [AIEEE-2002] (1) Absorbs CO2 and increase O2 concentration (2) Absorbs moisture (3) Absorbs CO2 (4) Produces ozone 3. In curing cement plasters, water is sprinkled from time to time. This helps in :- [AIEEE-2003] (1) Hydrating sand and gravel mixed with cement (2) Converting sand into silicate (3) Developing interlocking needle like crystals of hydrated silicates (4) Keeping it cool 4. The solubilities of carbonates decreases down the magnesium group due to decrease in- [AIEEE-2003] (1) Inter-ionic attraction (2) Entropy of solution formation (3) Lattice energy of solids (4) Hydration energy of cations 5. The substance not likely to contain CaCO3 is :- [AIEEE-2003] (1) Sea shells (2) Dolomite (3) A marble statue (4) Calcined gypsum 6. One mole of magnesium nitride on reaction with excess of water gives :- [AIEEE-2004] (1) Two mole of HNO3 (2) Two mole of NH3 (3) 1 mole of NH3 (4) 1 mole of HNO3 7. The ionic mobility of alkali metal ioins in aqueous solution is maximum for :- [AIEEE-2006] (1) Rb+ (2) Li+ (3) Na+ (4) K+ 8. Which of the following on thermal-decomposition yields a basic as well as an acidic oxide ? [AIEEE -20 12 ] (1) NH4NO3 (2) NaNO3 (3) KClO3 (4) CaCO3 9. Fire extinguishers contain H2SO4 and which one of the following :- [AIEEE-2012 (Online)] (1) CaCO3 (2) NaHCO3 and Na2CO3 (3) Na2CO3 (4) NaHCO3 10. Which one of the following will react most vigorously with water ? [AIEEE-2012 (Online)] (1) Li (2) K (3) Rb (4) Na 11. A metal M on heating in nitrogen gas gives Y. Y on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a blue colour, Y is :- [AIEEE-2012 (Online)] (1) NH3 (2) MgO (3) Mg3N2 (4) Mg(NO3)2 PREVIOUS YEAR QUESTIONS s-BLOCK EXERCISE-05(A) 9 10 11 Q. 1 2 3 4 5 6 7 8 333 A. 4 1 3 4 4 2 1 4
EXERCISE–05 (B) PREVIOUS YEARS QUESTIONS 1 . Which process is used in the extractive metallurgy of Mg : (A) Fused salt electrolysis (B) Self reduction (C) Aquaous solution electrolysis (D) Thermite reduction 2 . A sodium salt on treatment with MgCl2 gives white precipitate only on heating. The anion of sodium salt is : (A) H CO – (B) C O 2 – (C) N O – (D) SO 2 – 3 3 3 4 3 . The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order: (IIT 96) (I) K2CO3 (II) MgCO3 (III) CaCO3 (IV) BeCO3 (A) I < II < III < IV (B) IV < II < III < I (C) IV < II < I < III (D) II < IV < III < I 4 . Property of the alkaline earth metals that increases with their atomic number is – (IIT 97) (A) Ionisation energy (B) Solubility of their hydroxides (C) Solubility of their sulphates (D) Electronegativity 5 . The characteristics of solid sodium chloride are (REE 96) (1) Brittle (2) Ionic (3) Covalent (4) Non-conductor (A) 1 & 2 (B) 3 & 4 (C) 1, 2, & 4 (D) 1, 3, & 4 6 . Which of the following are not amphoteric – (REE 97) (1) Be(OH) (2) Sr(OH) (3) Ca(OH) (4) Al(OH) 2 2 2 3 (A) 1 & 3 (B) 2 & 3 (C) 1 & 4 (D) 2 & 4 7 . Highly dilute solution of sodium in liquid ammonia : (i) Shows blue colour (ii) Exhibits electrical conducitive (iii) Produces sodium amide (iv) Produces hydrogen gas (A) (i), (ii), (iii) (B) (i), (ii) (C) (iii), (iv) (D) Only (ii) 8 . Which of the following hydrides is not ionic (A) CaH2 (B) BaH2 (C) SrH2 (D) BeH2 9 . The compound(s) formed upon combustion of sodium metal in excess air is (are) [JEE 2009] (A) Na2O2 (B) Na2O (C) NaO2 (D) NaOH ASSERTION & REASON QUESTIONS (A) Statement–I is true, statement–II is true ; statement–II is a correct explanation for statement–I (B) Statement–I is true, statement–II is true ; statement–II is NOT a correct explanation for statement–I (C) Statement–I is true, statement–II is false (D) Statement–I is false, statement–II is true 1 . Statement–I : Alkali metals dissolve inz liquid ammonia to give blue solutions. (IIT 2007) Because : Statement–II : Alkali metals in liquid ammonia give solvated species of the type [M(NH )n]+ (M = alkali metals) 3
SUBJECTIVE QUESTIONS 1 . Identify the following: Na CO SO2 A Na2CO3 B Elemental S C I2 D 23 Also mention the oxidation state of S in all the compounds. 2 . Beryllium chloride shows acidic nature in water or why BeCl is easily hydrolysed? 2 3 . The crystalline salts of alkaline earth metals contain more water of crystallisation than the corresponding alkali metal salts, why? 4 . Arrange the following sulphates of alkaline earth metals in order of their decreasing thermal stability. BeSO4, MgSO4, CaSO4, SrSO4. 5 . Why the solubility of calcium acetate decreases while that of lead nitrate increases with increase in temperature. 6 . Why magnesium is not precipitated from a solution of its salt by NH OH In the presence of NH Cl. 44 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -05(B) 1. (A) 2. (A) 3. (B) 4. B 5. C 6. B 7. B 8. D 9. A, B Assertion - Reason Questions 1. B Subjective Questions 1 . A = NaHSO , Oxidation state of S = + 4 3 B = Na SO , Oxidation state of S = + 4 23 C = Na S O , Oxidation state of S = + 6 & – 2 22 3 D = Na S O , Oxidation state of S = + 5 & 0 24 6 2. BeCl is a salt of weak base Be(OH) and strong acid HCl and thus undergoes hydrolysis to result in an 22 acidic solution in water. BeCl + 4H O Hydration [Be(H O) ]2+ + 2Cl– 22 24 3. Alkaline earth metals have smaller size and more nuclear charge. 4. SrSO > CaSO > MgSO > BeSO 44 44 5. (CH COO) Ca shows exothermic dissolution whereas Pb(NO ) show endothermic dissolution. 32 32 6. The dissociation of NH4OH (a weak electrolyte) is suppressed in presence of NH4Cl due to common ion effect. Thus, [OH–] in solution becomes low. The ionic product of concentrations of Mg2+ and OH– ions does not exceed the solubility product of Mg(OH) and thus Mg(OH) is not precipitated. 22
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . A solid has a structure in which W atoms are located at the corners of a cubic lattice, O atom at the centre of the eges and Na atom at centre of the cubic. The formula for the compound is :- (A) NaWO2 (B) NaWO3 (C) Na2WO3 (D) NaWO4 2 . Which of the following statements is correct in the rock-salt structure of an ionic compounds? (A) coordination number of cation is four whereas that of anion is six. (B) coordination number of cation is six whereas that of anion is four. (C) coordination number of each cation and anion is four. (D) coordination number of each cation and anion is six. 3 . The coordination number of cation and anion in Fluorite CaF2 and CsCl are respectively :- (A) 8 : 4 and 6 : 3 (B) 6 : 3 and 4 : 4 (C) 8 : 4 and 8 : 8 (D) 4 : 2 and 2 : 4 4 . The interstitial hole is called tetrahedral because :- (A) it is formed by four spheres. (B) partly same and partly different (C) it is formed by four spheres the centres of which form a regular tetrahedron. (D) none of the above three 5 . The tetrahedral voids formed by ccp arrangement of Cl– ions in rock salt structure are :- (A) occupied by Na+ ions (B) occupied by Cl– ions (C) occupied by either Na+ or Cl– ions (D) vacant 6 . The number of nearest neighbours around each particle in a face-centred cubic lattice is :- (A) 4 (B) 6 (C) 8 (D) 12 7 . The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordi- nation number of eight. The crystal class is :- (A) simple cubic (B) body centred cubic (C) face centred cubic (D) none 8 . The mass of a unit cell of CsCl corresponds to :- (A) 1 Cs+ and 1 Cl– (B) 1 Cs+ and 6 Cl– (C) 4 Cs+ and 4 Cl– (D) 8 Cs+ and 1 Cl– 9 . In the closest packing of atoms A (radius : ra), the radius of atom B that can be fitted into tetrahedral voids is :- (A) 0.155 ra (B) 0.225 ra (C) 0.414 ra (D) 0.732 ra 1 0 . Which of the following will show schottky defect :- (A) CaF2 (B) Zns (C) AgCl (D) CaCl 1 1 . A cubic solid is made up of two elements A and B. Atoms B are at the corners of the cube and A at the body centre. What is the formula of compound. (A) AB(B) AB2 (C) A2B (D) None 1 2 . A compound alloy of gold and copper crystallizes in a cubic lattice in which gold occupy that lattice point at corners of the cube and copper atom occupy the centres of each of the cube faces. What is the formula of this compound. (A) AuCu6 (B) AuCu (C) AuCu3 (D) None
1 3 . KF crystallizes in the NaCl type structure. If the radius of K+ ions 132 pm and that of F– ion is 135 pm, what is the closet K–K distance ? (A) Can not say (B) 534 pm (C) 755 pm (D) 378 pm 1 4 . AgCl has the same structure as that of NaOH. The edge length of unit cell of AgCl is found to be 555 pm and the density of AgCl is 5.561 g cm–3. Find the percentage of sites that are unoccupied. (A) 0.24% (B) 2.4% (C) 24% (D) None 1 5 . The two ions A+ and B– have radii 88 and 200 pm respectively. In the closed packed crystal of compound AB, predict the co-ordination number of A+ . (A) 6 (B) 8 (C) 4 (D) None 1 6 . The effective radius of the iron atom is 1.42 Å. It has FCC structure. Calculate its density (Fe = 56 amu). (A) 2.87 gm/cm3 (B) 11.48 gm/cm3 (C) 1.435 gm/cm3 (D) 5.74 gm/cm3 CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B D C C D D B A B D A C D A A Que. 16 Ans. D
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . The density of CaF2 (fluorite structure) is 3.18 g/cm3. The length of the side of the unit cell is :- (A) 253 pm (B) 344 pm (C) 546 pm (D) 273 pm 2 . If the anions (A) form hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in it, then the general formula of the compound is :- (A) CA (B) CA2 (C) C2A3 (D) C3A2 3 . A solid is formed and it has three types of atoms X, Y, Z. X forms a FCC lattice with Y atoms occupying all the tetrahedral voids and Z atoms occupying half the octrahedral voids. The formula of the solid is :- (A) X2Y4Z (B) XY2Z4 (C) X4Y2Z (D) X4YZ2 4 . A compound XY crystallizes in BCC lattice with unit cell edge length of 480 pm. If the radius of Y– is 225 pm, then the radius of X+ is :- (A) 127.5 pm (B) 190.68 pm (C) 225 pm (D) 255 pm 5 . Which one of the following schemes of ordering closed packed sheets of equal sized spheres do not generate close packed lattice. (A) ABCABC (B) ABACABAC (C) ABBAABBA (D) ABCBCABCBC 6 . An ionic compound AB has ZnS type structure. If the radius A+ is 22.5 pm, then the ideal radius of B– would be :- (A) 54.35 pm (B) 100 pm. (C) 145.16 pm (D) none of these 7 . NH4Cl crystallizes in a body - centred cubic type lattice with a unit cell edge length of 387 pm. The distance between the oppositively charged ions in the lattice is :- (A) 335.1 pm (B) 83.77 pm (C) 274.46 pm (D) 137.23 pm 8 . rNa+ = 95 pm and rCl– = 181 pm in NaCl (rock salt) structure. What is the shortest distance between Na+ ions ? (A) 778.3 pm (B) 276 pm (C) 195.7 pm (D) 390.3 pm 9 . In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then radius of carbon atom is :- (A) 77.07 pm (B) 154.14 pm (C) 251.7 pm (D) 89 pm 1 0 . Give the correct order of initials T (true) or F (False) for following statements : I. In an anti-fluorite structure anions form FCC lattice and cations occupy all tetrahedral voids. II. If the radius of cations and anions are 0.2Å and 0.95Å then coordinate number of cation in the crystal is 4. III. An atom/ion is transferred from a lattice site to an interstitial position in Frenkel defect. IV. Density of crystal always increases due to substitutinal impurity defect. (A) TFFF (B) FTTF (C) TFFT (D) TFTF 1 1 . A cubic solid is made by atoms A forming close pack arrangement, B occupying one/fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of compound :- (A) A4B4C2 (B) A4B2C4 (C) A4BC (D) A4B2C2 1 2 . What is the percent by mass of titanium in rutile, a mineral that contain titanium and oxygen, if structure can be described as a closet packed array of oxide ions, with titanium in one half of the octahedral holes. What is the oxidation number of titanium ? (A) 30%, +4 (B) ~ 60%, +2 (C) ~ 60%, +4 (D) ~ 30%, +2
1 3 . Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one - eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by Zn2+, Al3+, O2–, with Zn2+ in the tetrahedral holes. Given the formula of spinel. (A) ZnAl2O4 (B) Zn2AlO4 (C) Zn2Al2O4 (D) None 1 4 . Calculate the density of diamond from the fact that it has face centered cubic structure with two atoms per lattice point and unit cell edge length of 3.569 Å. (A) 7 gm/cm3 (B) 1.75 gm/cm3 (C) 3.5 gm/cm3 (D) None 1 5 . An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit cell is 24 × 1024 cm3 and density of element is 7.2 g cm–3, calculate the number of atoms present in 200 g of element. (A) 1.1513 × 1024 (B) 3.472 × 1024 (C) 10.416 × 1024 (D) None 1 6 . Silver has an atomic radius of 144 pm and the density of silver is 10.6 cm–3. To which type of cubic crystal, silver belongs ? (A) HCP (B) BCC (C) Simple cubic (D) FCC 1 7 . Xenon crystallises in the face-centred cubic lattice and the edge of the unit cell is 620 pm. What is the next nearest neighbour distance ? (A) 738.5 pm (B) 620 pm (C) 438.5 pm (D) 310 pm 1 8 . CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl. (A) 141.42 pm (B) 282.84 pm (C) 346.4 pm (D) 173.2 pm 1 9 . An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Calculate the density, if 200 g of this element contains 5 × 1024 atoms. (A) 5 m/cm3 (B) 30 gm/cm3 (C) 10 gm/cm3 (D) 20 gm/cm3 2 0 . If the length of the body diagonal for CsCl which crystallises into a cubic structure with Cl– ions at the corners and Cs+ ions at the centre of the unit cells is 7 Å and the radius of the Cs+ ion is 1.69 Å. What is the radii for Cl– ion ? (A) 1.81 Å (B) 5.31 Å (C) 3.62 Å (D) none BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 Ans. B C A B C 6 7 8 9 10 11 12 13 14 15 Que. 16 17 18 19 20 Ans. D B C D A BADADDCACB
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . Crystalline solids are isotropic. 2 . Rhombohedral, triclinic and hexagonal are the unit cells, which have only primitive arrangement possible. 3 . Packing fraction of FCC and HCP units cells are same. 4 . The minimum void fraction for any unit cell in any shape having only one type of atom and all voids unfilled is 0.26. 5 . Packing fraction of a lattice structure depends on the radius of the atom crystallizing in it. 6 . The location of tetrahedral voids in FCC units cell are the centres of 8 minicubes forming a large cube. 7 . Effective number of octahedral voids in a unit cell is equal to the effective number of atoms in the unit cell. 8 . Radius ratio for co-ordination number 4 having tetrahedral and square planar geometry is same. 9 . The radius ratio value for co-ordination number 4 having square planar geometry and co-ordination number 6 having octahedral geometry is same. 1 0 . A metallic element crystallises into a lattice containing a sequence of layers AB AB AB ....... Any packing of spheres leaves out voids in the lattice. 26% percent by volume of this lattice is empty space. FILL IN THE BLANKS 1 . The relation between edge length (a) and radius of atom (r) for BCC lattice is ................. . 2 . The relation between edge length (a) and radius of atom (r) for FCC lattice is ................. . 3 . ABCABC ........ layering pattern is called ................. packing, found in ................. lattice. 4 . ABABAB........ layering pattern is called ................. packing, found in ................. lattice. 5 . Height (c) of the hexagonal primitive unit cell in term of radius of atom (r) is ................. . 6 . Anions would be in contact with each other only if the cation to anion radius for a given co-ordination number is ................. . 7 . The number of tetrahedral voids in hexagonal primitive unit cell is ................. . 8 . The limiting radius for co-ordination number 8 is ................. . 9 . For cesium chloride structure, the interionic distance (in terms of edge length, a) is equal to ................. . 1 0 . Density of a crystal ................. due to Schottky defect and ................. due to Frankel defect. MATCH THE COLUMN Column-II 1. Column-I (A) 68% occupy of space (p) Simple cubic lattice (B) CsCl (q) Diamond (C) Hexagonal close packing in three (r) Na O dimensions 2 (D) Antifluorite structure (s) AB AB type of close packing (E) Covalent crystal (t) Body centred cubic lattice.
2. Column-I Column-II (A) Spinel structure (p) Framework silicate (B) Glass (q) ZnFe O (C) Quartz (D) Metallic crystal 24 (E) Co-ordination number 6 (r) NaCl crystal (s) Pseudo solid (t) Melleable and ductile ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : Crystalline solids can cause X-rays to diffract. Because Statement-II : Interatomic distance in crystalline solids is of the order of 0.1 nm. 2 . Statement-I : Graphite is an example of tetragonal crystal system. Because Statement-II : For a tetragonal system a = b c, = = 90o, = 120o COMPREHENSION BASED QUESTIONS Comprehension # 1 Packing refers to the arrangement of constituent units in such a way that the forces of attraction among the constituent particles is maximum and the constituents occupy the maximum available space. In two dimensions, there are square close packing and hexagonal close packing. In three dimensions, however, there are hexagonal close packing, cubic close packing and body-centred cubic packing. (i) hcp : AB AB AB AB . . . arrangement Coordination no. = 12 % occupied space = 74 (ii) ccp : ABC ABC . . . arrangement Coordiantion no. = 12 % occupied space = 74 (iii) bcc : 68% space is occupied Coordination no. = 8 Answer the following questions : 1 . The empty space left in hcp in three dimensions is : (A) 26% (B) 74% (C) 52.4% (D) 80% 2 . In closed packed lattice containing 'n' particles, the numbers of tetrahedral and octahedral voids are : (A) n, 2n (B) n, n (C) 2n, n (D) 2n, n/2 3 . The pattern of successive layers of ccp arrangement can be designated as : (A) AB AB AB . . . (B) AB ABC AB ABC . . . (C) ABC ABC ABC . . . (D) AB BA AB BA . . . 4 . The space occupied by spheres in bcc arrangement is : (A) 74% (B) 70% (C) 68% (D) 60.4%
5 . A certain oxide of metal M crystallises in such a way that O2– ions occupy hcp arrangement following AB 2 AB . . . pattern. The metal ions, however, occupy 3 rd of the octahedral voids. The formula of the com- pound is (A) M O (B) M O (C) M O (D) MO 23 3 8/3 3 2 Comprehension # 2 A2+ B3+ O2+ Answer the following questions : 1 . The space lattice given in the figure refers to : (A) fluoride structure (B) rock salt structure (C) spinel structure (D) inverse spinel structure 2 . O2– ions are present in : (A) bcc arrangement (B) fcc arrangement (C) simple cubic arrangement (D) hcp arrangement 3 . The formula of the compound is : (A) ABO (B) A BO (C) AB O (D) A BO 2 23 24 24 4 . Fraction of the total octahedral voids occupied will be : 11 1 1 (A) (B) (C) 8 (D) 6 24 5 . B3+ and A2+ ions are present in : (A) tetrahedral voids (B) octahedral, tetrahedral voids (C) tetrahedral, octahedral voids (D) octahedral cubic voids MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 7. T 1. F 2. T 3. T 4. T 5. F 6. T 10. T 8. F 9. T Fill in the Blanks 1. 3a 4r 2. 2a 4r 3. cubic close, FCC 4. Hexagonal close, HCP 5. c 4r 2 6. least or minimum 7. 12 8. 0.732 3 3a 10. decreases, remains constant 9. 2 Match the Column 1. (A) t; (B) p ; (C) s ; (D) r ; (E) q 2. (A) q; (B) s ; (C) p ; (D) t ; (E) r Assertion - Reason Questions 1. C 2. D Comprehension Based Questions Comprehension # 1 : 1. (A) 2. (C) 3. (C) 4. (C) 5. (A) Comprehension #2 : 1. (C) 2. (B) 3. (C) 4. (A) 5. (B)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . A closed packed structure of uniform spheres has the edge length of 534 pm. Calculate the radius of sphere, if it exist in :- (a) simple cubic lattice (b) BCC lattice (c) FCC lattice 2 . Gold crystallizes in a face centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu:- 3 . The density of KBr is 2.75 g cm–3. The length of the edge of the unit cell is 654 pm. Show the KBr has face centered cubic structure. 4 . A crystal of lead (II) sulphide has NaCl structure. In this crystal the shortest distance between Pb+2 ion and S2– ions is 297 pm. What is the length of the edge of the unit cell in lead sulphide ? Also calculate the unit cell volume. 5 . Iron has body centred cubic lattice structure. The edge length of the unit cell is found to be 286 pm. What is the radius of an iron atom ? 6 . Cesium chloride forms a body centred cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is 412 pm and Cl– ion has a radius of 181 pm. Calculate the radius of Cs+ ion. 7 . In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eight of tetrahedral voids are occupied by divalent ions (A2+) while one half of the octahedral voids occupied trivalent ions (B3+). What is the formula of the oxide ? 8 . A solid A+ and B– has NaCl type closed packed structure. If the anion has a radius of 250 pm, what should be the ideal radius of the cation ? Can a cation C+ having a radius of 180 pm be slipped into the tetrahedral site of the crystal of A+B– ? Give reasons for your answer. 9 . Calculate the value of Avogadro's number from the following data : Density of NaCl = 2.165 cm–3 Distance between Na+ and Cl– in NaCl = 281 pm. 1 0 . If the radius of Mg2+ ion, Cs+ ion, O2– ion, S2– ion and Cl– ion are 0.65 Å, 1.69 Å, 1.40 Å, 1.84 Å, and 1.81Å respectively. Calculate the co-ordination numbers of the cations in the crystals of MgS, MgO and CsCl. 1 1 . Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is 124 pm. compute the density of iron in both these structures. 1 2 . KCl crystallizes in the same type of lattice as does NaCl. Given that rNa = 0.5 and rNa = 0.7. Calculate : r rCl K (a) The ratio of the sides of unit cell for KCl to that for NaCl and (b) The ratio of densities of NaCl to that for KCl. 1 3 . An element A(Atomic weight = 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A. 1 4 . Prove that the void space percentage in zinc blende structure is 25%. 1 5 . A unit cell of sodium chloride has four formula units. The edge of length of the unit cell is 0.564 nm. What is the density of sodium chloride. 1 6 . In a cubic crystal of CsCl (density = 3.97 gm/cm3) the eight corner are occupied by Cl– ions with Cs+ ions at the centre. Calculate the distance between the neighbouring Cs+ and Cl– ions. 1 7 . The composition of a sample of wustite is Fe0.93O1.0. What percentage of iron is present in the form of Fe(II)? 1 8 . Rbl crystallizes in bcc structure in which each Rb+ is surrounded by eight iodide ions each of radius 2.17 Å.
Find the length of one side of RbI unit cell. 1 9 . If NaCl is dopped with 10–3 mol % SrCl2, what is the number of cation vacancies? 2 0 . NaH crystallizes in the same structure as that NaCl. The edge length of the cubic unit cell of NaH is 4.88 Å. (a) Calculate the ionic radius of H–, provided the ionic radius of Na+ is 0.95 Å. (b) Calculate the density of NaH. 2 1 . Ice crystallizes in a hexagonal lattice. At the low temperature at which the structure was determined, the lattice constants were a = 4.53Å, and b = 7.60Å (see figure). How many molecules are contained in a given unit cell? [density (ice) = 0.92 gm/cm3] b a 2 2 . Using the data given below, find the type of cubic lattice to which the crystal belongs. Fe V Pb a in pm 2 8 6 3 0 1 3 8 8 in gm cm–3 7.86 5.96 12.16 2 3 . Prove that void space in fluoride structure per unit volume of unit cell is 0.243. 2 4 . A compound formed by elements X & Y. Crystallizes in a cubic structure, where X is the at corners of the cube and Y is at six face centres. What is the formula of the compound ? If side length is 5Å, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively. 2 5 . The metal nickel crystallizes in a face centred cubic structure. Its density is 8.9 gm/cm3. Calculate (a) the length of the edge of the unit cell. (b) the radius of the nickel atom. [Atomic weight of Ni = 58.89] CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1. 267 pm, 231.2 pm, 188.8 pm 2. 19.4 g/cm3 , 143.9 pm 3. N = 6.023 × 1023 mol–1, At. mass : K = 39, Br = 80 4. a = 5.94 × 10–8 cm, V = 2.096 × 10–22 cm–3 5. 123.84 pm 6. 175.8 pm 7. AB2O4 8. 103.4 pm, No 9. 6.01 × 1023 10. 4, 6, 8 11. 7.887 g/cc, 8.59 gm/cm3 12. (a) 1.143, (b) 1.172 13. 5.188 gm/cm3, 6.023 × 1022 atoms of A, 3.0115 × 1022 unit cells 15. 2.16 gm/cm3 16. 3.57 Å 17. 15.053 18. 4.34 Å 19. 6.02 × 1018 mol–1 20. (a) 1.49 Å, (b) 1.37 g/cm3 21. 4 molecules of H O 22. for Fe is bcc, for V is bcc, for Pd is face centred 2 25. (a) 3.52 Å, (b) 1.24 Å 24. XY , 4.38 g/cm3 3
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . The element chromium exists as a bcc lattice whose unit cell edge is 2.88 Å. The density of chromium is 7.20 g/cc. How many atom does 52.0 g of chromium contain ? 2 . The edge length of the unit cell of KCl (NaCl like structure ; fcc) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. (rCl– = 1.8173 Å). 3 . A cubic unit cell contains manganese ions at the corners and fluoride ions at the centre of each edge. (a) What is the empirical formula ? (b) What is the C.N. of the Mn ion? (c) Calculate the edge length of the unit cell if the radius of a Mn ions is 0.65 Å and that of F– ion is 1.36Å. (d) Calculate the density of the compound (Mn = 55, F = 19). 4 . Silver crystallises in fcc lattice. If edge length of the cell is 4.077 × 10–8 cm and density is 10.5 g cm–3, calculate the atomic mass of silver. 5 . Formula mass of NaCl is 58.45 g mol–1 and density of its pure form is 2.167 g cm–3. The average distance between adjacent sodium and chloride ions in the crystal is 2.814 × 10–8 cm. Calculate Avogadros constant. 6 . Thallium chloride, TlCl (240 g mol–1) crystallises in either a simple cubic lattice or a face centered cubic lattice Cl– ions with Ti+ ions in the holes. If the density of the solid is 9.00 g cm–3 and edge of the unit cell is 3.85 × 10–8 cm, what is the unit geometry ? 7 . KF has NaCl structure. What is the distance between K+ and F– in KF, if the density is 2.48 g cm–3 ? 8 . BaTiO crystallizes in the prevoskite structure. The structure may be described as a cubic lattice with barium 3 ions occupying the corner of the unit cell, oxide ions occupying the face-centers and titanium ion occupying the center of the unit cell. (a) If titanium is described as occupying holes in BaO lattice, what type of holes does it occupy ? (b) What fraction of this type hole does it occupy ? 9 . Find the size of largest sphere that will fit in octahedral void in an ideal FCC crystal as a function of atomic radius 'r'. The insertion of this sphere into void does not distort the FCC lattice. Calculate the packing fraction of FCC lattice when all the octahedral voids are filled by this sphere. 1 0 . Metallic gold crystallises in fcc lattice. The length of the cubic unit cell is a = 4.07 Å. (a) What is the closest distance between gold atoms. (b) How many \"nearest neighbours\" does each gold atom have at the distance calculated in (a). (c) What is the density of gold? (d) Prove that the packing fraction of gold is 0.74. 1 1 . Potassium crystallizes in a body-centered cubic lattice with edge length, a = 5.2 Å. (a) What is the distance between nearest neighbours? (b) What is the distance between next-nearest neighbours? (c) How many nearest neighbours does each K atom have? (d) How many next-neighbours does each K atom have? (e) What is the calculated density of crystalline potassium? 1 2 . The olivine series of minerals consists of crystals in which Fe and Mg ions many substitute for each other causing substitutional impurity defect without changing the volume of the unit cell. In olivine series of minerals, oxide ion exist as FCC with Si4+ occupying 1 1 th of octahedral voids and divalent ions occupying th of 44 tetrahedral voids. The density of forsterite (magnesium silicate) is 3.21 g/cc and that of fayalite (ferrous silicate) is 4.43 g/cc. Find the formula of forsterite and fayalite minerals and the percentage of fayalite in an olivine with a density of 3.88 g/cc.
13 . The mineral hawleyite, one form of CdS, crystallizes in one of the cubic lattices, with edge length 5.87 Å. The density of hawleyite is 4.63 g cm–3. (i) In which cubic lattice does hawleyite crystallize? (ii) Find the Schottky defect in g cm–3. 1 4 . A strong current of trivalent gaseous boron passed through a germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium is taken and the boron atoms are found to be 150 ppm by weight, when the density of the Ge crystal decreases by 4%. Calculate the percentage of missing vacancies due to germanium, which are filled up by boron atoms. Atomic wt. Ge = 72.6, B = 11 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1. 6.05 × 1023 atoms mol–1 2. rk+ = 1.3227 Å 11 (b) 6(fcc) (c) 4.02 Å (d) 2.86 g/cm3 3. (a) MnF (Mn = 8 × ; F = 6 × ) 5. 6.05 × 1023 mol–1 382 4. 107.09 g mol–1 6. simple cubic battice 7. 269 pm 8. (a) octahedral (b) 1/4 10. (a) 2.88 Å (b) 12 (c) 19.4 g/cc 9. 0.414 r, 79.3% 6 (e) 0.92 g/cm3 11. (a) 4.5 Å (b) 5.2 Å (c) 8 (d) 12. Mg SiO , Fe SiO , 59% 13. (i) 3.90 (ii) 0.120 g/cc 24 24 14. 2.376 %
EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS Q.1 Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic Q.2 Q.3 radius of the lithium will be :- [AIEEE-2012] Q.4 Q.5 (A) 152 pm (B) 75 pm (C) 300 pm (D) 240 pm Q.6 Q.7 In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face Q.8 Q.9 centre positions. If one atom of B is missing from one of the face centred points, the formula of the Q.10 compound is :- [AIEEE-2011] Q.11 Q.12 (A) A2B3 (B) A2B5 (C) A2B (D) AB2 Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is:- [AIEEE-2011] (A) 181 pm (B) 108 pm (C) 128 pm (D) 157 pm Percentages of free space in cubic close packed structure and in body centered packed structure are respectively :- [AIEEE-2010] (A) 48% and 26% (B) 30% and 26% (C) 26% and 32% (D) 32% and 48% The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is :- [AIEEE-2010] (A) 144 pm (B) 288 pm (C) 398 pm (D) 618 pm In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be - [AIEEE-2008] (A) X4Y3 (B) X2Y3 (C) X2Y (D) X3Y4 Total volume of atoms present in a face-centred cubic unit cell of a metal is (r is atomic radius) : [AIEEE-2006] (A) 24 r3 (B) 12 r3 (C) 16 r3 (D) 20 r3 3 3 3 3 Lattice energy of an ionic compound depends upon - [AIEEE-2005] (A) Size of the ion only (B) Charge on the ion only (C) Charge on the ion and size of the ion (D) Packing of ions only An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula of this compound would be-[AIEEE-2005] (A) A2B (B) AB (C) A3B (D) AB3 [AIEEE-2004] What type of crystal defect is indicated in the diagram below ? Na+ Cl– Na+ Cl– Na+ Cl– Cl– Cl– Na+ Na+ Na+ Cl– Cl– Na+ Cl– Cl– Na +Cl– Na+ Na+ (A) Frenkel defect (B) Schottky defect (C) Interstitial defect (D) Frenkel and Schottky defects How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00g ? (A) 1.28 × 1021 unit cells (B) 1.71 × 1021 unit cells [AIEEE-2003] (C) 2.57 × 1021 unit cells (D) 5.14 × 1021 unit cells The no. of atoms per unit cell in B.C.C. & F.C.C. is respectively : [AIEEE-2002] (A) 8, 10 (B) 2, 4 (C) 1, 2 (D) 1, 3
JEE-[MAIN] : PREVIOUS YEAR QUESTIONS ANSWER KEY EXERCISE -5[A] Q.3 (C) Q.1 (A) Q.2 (B) Q.7 (C) Q.4 (C) Q.11 (C) Q.8 (C) Q.5 (A) Q.6 (A) Q.12 (B) Q.9 (D) Q.10 (B)
EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS Q.1 A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is : [JEE-2012] M= X= (A) MX (B) MX2 (C) M2X (D) M5X14 Q.2 The number of hexagonal faces that present in a truncated octahedron is. [JEE-2011] Q.3 The packing effeciency of the two-dimensional square unit cell shown below is [JEE-2010] (A) 39.27% (2) 68.02% (C) 74.05% (D) 78.54% Q.4 The correct statement(s) regarding defects in solid is (are) [JEE 2009] (A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion. (B) Frenkel defect is a dislocation defect (C) Trapping of an electron in the lattice leads to the formation of F-center. (D) Schottky defects have no effect on the physical properties of solids. Paragraph for Question No. 5 to 7 In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ‘r’. Q.5 The number of atoms in this HCP unit cells is [JEE 2008] (A) 4 (B) 6 (C) 12 (D) 17
Q.6 The volume of this HCP unit cell is [JEE 2008] (A) 24 2 r3 (B) 16 2 r3 (C) 12 2 r3 (D) 64 r3 33 Q.7 The empty space in this HCP unit cell is [JEE 2008] Q.8 (A) 74% (B) 47.6 % (C) 32% (D) 26% Q.9 Q.10 Match the crystal system / unit cells mentioned in Column I with their characteristic features mentioned in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I Column II (A) simple cubic and face-centred cubic (P) have these cell parameters a = b = c and = = (B) cubic and rhombohedral (Q) are two crystal systems (C) cubic and tetragonal (R) have onlytwo crystallographic angles of 90° (D) hexagonal and monoclinic (S) belong to same crystal system. [JEE 2007] The edge length of unit cell of a metal having atomic weight 75 g/mol is 5 Å which crystallizes in cubic lattice. If the density is 2 g/cc then find the radius of metal atom. (NA = 6 × 1023). Give the answer in pm. [JEE 2006] An element crystallises in FCC lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interstital sites without disturbing the structure. [JEE 2005] Q.11 Which of the following FCC structure contains cations in alternate tetrahedral voids? (A) NaCl (B) ZnS (C) Na2O (D) CaF2 [JEE 2005] Q.12 (i) AB crystallizes in a rock salt structure with A : B = 1 : 1. The shortest distance between A and B is Y1/3 nm. The formula mass of AB is 6.023 Y amu where Y is any arbitrary constant. Find the density in kg m-3. (ii) If measured density is 20 kg m-3. Identify the type of point defect. [JEE–2004] Q.13 Marbles of diameter 10 mm each are to be arranged on a flat surface so that their centres lie within the area enclosed by four lines of length each 40 mm. Sketch the arrangement that will give the maximum number of marbles per unit area, that can be enclosed in this manner and deduce the expression to calculate it. [JEE 2003] Q.14 A substance AxBy crystallises in a FCC lattice in which atoms “A” occupy each corner of the cube and atoms “B” occupy the centres of each face of the cube. Identify the correct composition of the substance AxBy. (A) AB3 (B) A4B3 (C) A3B (D) composition cannot be specified [JEE-2002]
Q.15 The figures given below show the location of atoms in three crystallographic planes in FCC lattice. Draw the unit cell for the corresponding structure and identify these planes in your diagram. [JEE-2000] Q.16 In a solid “AB” having NaCl structure “A” atoms occupy the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is (A) AB (B) A B (C) A B (D) A B [JEE-2000] 2 2 43 34 Q.17 In any ionic solid [MX] with schottky defects , the number of positive and negative ions are same. [T/F] [JEE-2000] Q.18 The coordination number of a metal crystallising in a hcp structure is [JEE-2000] (A) 12 (B) 4 (C) 8 (D) 6 Q.19 A metal cryatallises into two cubic phases, FCC and BCC whose unit cell lengths are 3.5 and 3.0 Å respectively. Calculate the ratio of densities of FCC and BCC. [JEE-1999] PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5[B] Q.1 (B) Q.2 (8) Q.3 (D) Q.4 (B,C) Q.5 (B) Q.6 (A) Q.7 (D) Q.8 (A) P, S ; (B) -P,Q ; (C) - Q ; (D) - Q,R Q.9 216.5 pm Q.10 117.1 pm Q.11 (B) Q.12 (i) = 5 kg m–3 (ii) There is huge difference in theoretically calculated density and observed density. It is only possible if some foreign species occupies interstitial space i.e. substitution defect. Q.13 Q.14 (A) Q.15 Q.16 (D) Q.17 True Q.18 (A) Q.19 1.259
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The molarity of a glucose solution containing 36 g of glucose per 400 mL of the solution is: (A) 1.0 (B) 0.5 (C) 2.0 (4) 0.05 2 . 1 kg of NaOH solution contains 4g of NaOH. The approximate concentration of the solution is: (A) 0.1 molar (B) 0.1 molal (C) Decinormal (D) About 0.1 N 3 . To prepare 0.1 M KMnO solution in 250 mL flask, the weight of KMnO required is: 44 (A) 4.80g (B) 3.95g (C) 39.5g (D) 0.48 g 4 . The number of moles present in 2 litre of 0.5 M NaOH is: (A) 2 (B) 1 (C) 0.1 (D) 0.5 5 . The weight of solute present in 200 mL of 0.1 M H2SO4 : (A) 2.45g (B) 4.9g (C) 1.96g (D) 3.92 g 6 . The nature of mixture obtained by mixing 50 mL of 0.1 M H SO and 50 mL of 0.1 M NaOH is: 24 (A) Acidic (B) Basic (C) Neutral (D) Amphoteric 7 . If 250 mL of a solution contains 24.5g H2SO4 the molarity and normality respectively are: (A) 1M, 2N (B) 1M,0.5M (C) 0.5M, 1N (D) 2M,1N 8 . The volume strength of H2O2 solution is 10. What does it mean : (A) at S.T.P. 10 g solution of H2O2 gives 10 mL of O2 (B) at S.T.P. 1 g equivalent of H O gives 10 mL of O 2 22 (C) at ST.P. 10 litre solution of H2O2 gives 10 mL of O2 (D) at S.T.P. 1 mL solution of H2O2 gives 10 mL of O2 9 . The normality of 0.3 M phosphorus acid (H3PO3) is: (A) 0.1 (B) 0.9 (C) 0.3 (D) 0.6 1 0 . The normality of 4% (wt./vol.) NaOH is: (A) 0.1 (B)1.0 (C) 0.05 (D) 0.01 1 1 . The density of NH4OH solution is 0.6 g/mL. It contains 34% by weight of NH4OH. Calculate the normality of the solution : (A) 4.8 N (B) 10 N (C) 0.5 N (D) 5.8 N 1 2 . A molal solution is one that contains one mole of a solute in : (A) 1000 g of the solution (B) 1000 c.c. of the solution (C) 1000 c.c of the solvent(D) 1000 g of the solvent 1 3 . Out of molarity (M), molality (m), formality (F) and mole fraction (x) those independent of temperature are : (A) M,m (B) F, x (C) m,x (D) M,x 1 4 . 3.0 molal NaOH solution has a density of 1.110 g/mL. The molarity of the solution is: (A) 2.9732 (B) 3.05 (C) 3.64 (D) 3.0504 1 5 . 1000 gram aqueous solution of CaCO3 contains 10 gram of carbonate. Concentration of solution is: (A) 10ppm (B) 100ppm (C) 1000ppm (D) 10,000 ppm 1 6 . When 5.0 gram of BaCl is dissolved in water to have 106 gram of solution. The concentration of solution is 2 : (A) 2.5 ppm (B) 5 ppm (C) 5M (D) 5 g L–1 1 7 . How many grams of glucose be dissolved to make one litre solution of 10% glucose : (A) 10g (B) 180g (C) 100g (D) 1.8g
1 8 . Vapour pressure of a solvent containing nonvolatile solute is : (A) more than the vapour pressure of a solvent (B) less than the vapour pressure of solvent (C) equal to the vapour pressure of solvent (D) none 1 9 . The relative lowering in vapour pressure is: (A) X (B) 1 (C) = X (D) m solute X solute solute 2 0 . The vapour pressure of a dilute solution of a solute is not influenced by : (A) temperature of solution (B) melting point of solute (C) mole fraction of solute (D) degree of dissociation of solute 2 1 . An aqueous solution of methanol in water has vapour pressure : (A) equal to that of water (B) equal to that of methanol (C) more than that of water (D) less than that of water 2 2 . When a substance is dissolved in a solvent, the vapour pressure of solvent decreases. This brings: (A) an increase in b.pt. of the solution (B) a decrease in b.pt of a solution (C) an increase in f.pt of the solvent (D) none 2 3 . Solute when dissolved in water: (A) increases the vapour pressure of water (B) decreases the boiling point of water (C) decreases the freezing point of water (D) all of the above 2 4 . If the vapour pressure of solutions of two liquids are less than those expected from ideal solution they are said to have : (A) negative deviation from ideal behaviour (B) positive deviations from ideal behaviour (C) ideal behaviour (D) positive deviation for lower concentration and negative deviations for higher concentration 2 5 . A 5.8% solution of NaCl has vapour pressure closest to : (A) 5.8 % solution of urea (B) 2 m solution of glucose (C) 1 m solution of urea (D) 5.8 % solution of glucose 2 6 . The boiling point of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C respectively. Which will show highest vapour pressure at room temperature: (A) C6H6 (B) CH3OH (C) C6H5NH2 (D) C6H5NO2 2 7 . Boiling point of water is defined as the temperature at which : (A) vapour pressure of water equal to that of atmospheric pressure (B) bubbles are formed (C) steam comes out (D) none of the above 2 8 . Which solution will show maximum elevation in b.pt: (A) 0.1 M KCl (B) 0.1 M BaCl2 (C) 0.1 M FeCl3 (D) 0.1 M Fe2(SO4)3 2 9 . The correct relationship between the boiling points of very dilute solutions of AICI3 (t1) and CaCI2(t2) having the same molar concentration is : (A) t1 = t2 (B) t1 > t2 (C) t2 > t1 (D) t2 t1 3 0 . Cryoscopic constant of a liquid is: (A) decrease in freezing point when 1 gram of solute is dissolved per kg of the solvent (B) decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent (C) the elevation for 1 molar solution (D) a factor used for calculation of elevation in boiling point
3 1 . At certain Hill-station pure water boils at 99.725°C. If K for water is 0.513°C kg mol–1, the boiling point of b 0.69 m solution of urea will be : (A) 100.079°C (B) 103°C (C) 100. 359°C (D) un predictable 3 2 . The freezing point of 1 molal NaCl solution assuming NaCI to be 100% dissociated in water is : (A) –1.86°C (B) –3.72°C (C) +1.86°C (D) +3.72°C 3 3 . 10 gram of solute with molecular mass 100 gram mol–1 is dissolved in 100 gram solvent to show 0.3°C elevation in boiling point. The value of molal ebullioscopic constant will be : (A) 10 (B) 3 (C) 0.3 (D) un predictable 3 4 . Depression in freezing point of solution of electrolytes are generally: (A) lower (B) higher than what should be normally (C) low or high depending upon nature of electrolyte (D) what it should be normally 3 5 . A liquid is in equilibrium with its vapour at its boiling point. On the average the molecules in the two phase have equal: (A) inter-molecular forces (B) potential energy (C) total energy (D) kinetic energy 3 6 . Which salt may show the same value of vant Hoff factor (i) as that of K4Fe(CN)6 in very dilute solution state : (A) Al2(SO4)3 (B) NaCl (C) Al(NO3)3 (D) Na2SO4 3 7 . Which compound corresponds vant Hoff factor (i) to be equal to 2 in dilute solution: (A) K2SO4 (B) NaHSO4 (C) Sugar (D) MgSO4 3 8 . In which of the following, the vant Hoff factor (i) is equal to one: (A) NaCl (B) KNO3 (C) Urea (D) all 3 9 . If the observed and theoretical molecular mass of NaCl is found to be 31.80 and 58.50, then the degree of dissociation of NaCl is : (A) 83.96% (B) 8.39% (C) 90% (D) 100% 4 0 . The substance A when dissolved in solvent B shows the molecular mass corresponding to A3. The vant Hoffs factor will be: (A) 1 (B) 2 (C) 3 (D) 1/3 4 1 . Which of the following conditions is not correct for ideal solution : (A) no change in volume on mixing (B) no change in enthalpy on mixing (C) it obey’s Raoult’s law (D) lonisation of solute should occurs to a small extent 4 2 . Solutions distilled without change in composition at a temperature are called : (A) Amorphous (B) Azeotropic mixture (C) Ideal solution (D) Super saturated solution 4 3 . If mole fraction of the solvent in a solution decreases then : (A) vapour pressure of solution increases (B) b.pt decreases (C) osmotic pressure increases (D) all are correct 4 4 . An azeotropic solution of two liquids has boiling point lower than either of them when it: (A) shows a negative deviation from Raoult’s law (B) shows no deviation from Raoult’s law (C) shows positive deviation from Raoult’s law (D) is saturated 4 5 . The passing of particles through semipermeable membrane is called : (A) osmosis (B) electrodialysis (C) electrophrosis (D) electroplating
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