Tactics of Passing Chemistry

Learning chemistry made easier ii. Draw a well labeled diagram to show how a copper spoon can be electroplated with silver. (2 marks) Copper spoon Ag+(aq) iii. During the electrolysis of aqueous magnesium sulphate, a current of 0.5A was passed for 21 minutes and 36 seconds. Calculate the volume of gas produced at the anode. (1 Farady = 96,500C, Molar gas volume = 24 dm3)(3 marks) Quantity of electricity passed = Current x time = 0.5 x (21 x 60 + 36) = 0.5 x 1, 296 = 648 Coulombs For one mole of oxygen gas to be liberated 4 moles of electrons are releases according to the equation. 4OH-(aq) 2H2O(l) + O2(g) + 4e- The charge = 4 x 96, 500 = 386, 000 Coulombs The volume of oxygen produced by 648C is = 648 x 24 = 0.04029 dm3 or 40.29cm3. 386, 000 Other binary electrolyte questions requires you to know which ions goes where. For instance when electrolyzing molten NaCl salt, Na+ migrate to Cathode and Cl- to anode. 89

Tactics of Chemistry Tactical Practise K 1. Draw a set up that can be used to separate a mixture of sand and iodine. (3 marks) 2. Draw a labelled diagram to illustrate how alpha, beta and gamma radiations can be distinguished from each other. ( 3 marks) 3. In an experiment, dry hydrogen gas was passed over hot copper (II) oxide in a combustion tube as shown in the diagram below:- Copper (II) Oxide Substance R DHryydrogen gas Heat Complete the diagram to show how the other product, substance R could be collected in the laboratory. (3 marks) 4. Maalim set-up the experiment as shown below to collect a gas. The wet sand was heated before heating Zinc granules Zinc Wet sand Complete the diagram for the laboratory preparation of the gas 5. The set-up below was used to investigate electrolysis of a certain molten compound;- (3 marks) I- Pb2+ Molten substance a. Complete the circuit by drawing the cell in the gap left in the diagram b. Write half-cell equation to show what happens at the cathode c. Using an arrow show the direction of electron flow in the diagram above 6. The set-up below is used to prepare dry sulphur (IV) Oxide in the laboratory. Answer questions that follow: 90

Learning chemistry made easier Dilute H2SO4 Gas jar Sodium Sulphite Concentrated H2SO4 a. Identify the mistake in the set-up (1 mark) b. Write an equation for the reaction in the set-up. ( 1 mark) c. State how the polluting effects of the gas on the environment can be controlled. ( 1 mark) 7. Study the standard electrode potential for the half-cells given below and answer the questions that follow. (The letter do not represent the actual symbols of the elements) N+ +e- N(s) ; -2.92V (aq) J(s) ; +0.52V J+ + e- (aq) ½ K2(g) ; 0.00V K+ + e- (aq) ½ G2(g) + e- ; +1.36V G- (aq) M(s) ; -0.44V M2+ + 2e- (aq) i. Identify the strongest oxidizing agents. Give a reason for your answer. (2 marks) ii. What two half-cells would produce the highest potential difference when combined? (1 mark) iii. Draw a complete electrochemical cell of the two-half cells mentioned in (ii) above. (3 marks) 8. Draw the diagrams for the laboratory preparation of the following gases. • Dry Oxygen gas • Dry nitrogen (I) oxide • Dry hydrogen gas • Dry nitrogen (II) oxide • Dry carbon (II) oxide • Dry nitrogen (IV) oxide • Dry carbon (IV) oxide • Dry ammonia gas • Dry methane • Dry sulphur (IV) oxide • Dry ethane • Dry Sulphur (VI) oxide • Dry ethene • Dry hydrogen sulphide gas • Dry ethyne • Dry chlorine gas • Dry nitrogen • Dry hydrogen chloride gas 91

Tactics of Chemistry 4. CALCULATIONS & DETERMINATIONS Calculations are always marked consequentially in practical paper only. In case a step is wrong, you are penalized at that step or point only and the rest is right if the right procedure is followed. Let us look at the questions below. Sample question 55 You are provided with:- Solution A, 0.07M hydrochloric acid 1g solid B, Calcium hydroxide You are required to determine the solubility of Ca (OH)2 Procedure: Transfer all solid B into a 250cm3 volumetric flask. Measure accurately using a clean measuring cylinder 50 cm3 of water and transfer this carefully into the volumetric flask. Shake gently and measure a second portion of 40cm3 water and add this to the resulting solution in the volumetric flask. Filter the solution into a beaker and label this solution D. Place solution A in the burette, pipette 25.0 cm3 of solution D into a 250 cm3 conical flask and titrate using methyl orange indicator. Record your result in table below and repeat the titration carefully to obtain consistent results. Titration I II III Final Burette reading (cm3) 18.50 39.00 19.50 Initial Burette reading (cm3) 0.00 19.50 0.00 (4marks) Volume of A used (cm3) 19.50 19.50 19.50 Calculate: a. Volume of solution A used. ( 1mark) = 19.50 + 19.50 + 19.50 = 19.50cm3 3 b. Number of moles of the solution A reacted. ( 1mark) 92

Learning chemistry made easier Student I Student II mv 0.07 x 25 mv 0.07 x 19.5 1000 1000 1000 1000 = 0.00175 moles = 0.001365 moles Student I used the wrong volume of 25cm3 instead of the correct volume as in student II of 19.5cm3. Student I does not score this step but student II scores full marks. c. Number of moles of solution D in the 25cm3 . (2marks) Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(aq) Ratio 1:2 Student II Student I Moles = 0.001365 Moles = 0.00175 2 2 = 0.000875 moles = 0.0006825 moles Both students scores full marks. d. Calculate mole of solution D in the 90 cm3 of the solution D. (1mark) Student I Student II 0.000875 = 25 cm3 0.0006825 = 25 cm3 90cm3 90cm3 0.000875 x 90 = 0.00315 mol 0.0006828 x 90 = 0.002457 mol 25 25 Both students scores full marks. 93

Tactics of Chemistry e. Calculate the mass of calcium hydroxide that dissolved in 90cm3 of water. (2marks) Student I Student II Mass = 0.00315 x 74 Mass = 0.002457 x 74 = 0.2331g = 0.181g Both students scores full marks. f. Determine the solubility of calcium hydroxide at the room temperature. (2marks) Student I Student II 0.2331g contained in 90cm3 0.181g contained in 90cm3 100cm3 = 0.2331 x 1000 100cm3 = 0.181 x 1000 90 90 = 0.259g in 100g H2O = 0.2g in 100g H2O Student II will score all the marks in the final answer while student I will be penalized 1 mark for the final answer. It is not the correct answer. The Tactic If you get a wrong value at a particular level, but use the value well in subsequent calculations you will score, all points as allocated in subsequent steps though the final answer may also be exempted. Other common mistakes that students forget or they don`t know is the correct relative formula/molecular mass. A student for instance may miss to use the relative molecular mass of cSaOl2cualsat6e4d. instead use 32, but the other subsequent steps are well Such a student will be penalized for that wrong step missed in calculating the relative molecular mass of SO2 and the final answer but scores all 94

Learning chemistry made easier the other steps if well calculated. Avoid calculating beyond the limit or manipulating your calculation to fit your answer. Sample question 56 How many molecules are there in 20g of chlorine gas. (Cl = 35.5, Avogadro = 6.0 x 1023. Moles of Cl2 = 20 = 0.2817 x 6.0 x 1023 71 = 1.6902 x 1023 = 0.2817 moles = 1.6902 x 1023 x 2 = 3.3804 x 1023 Number of molecules The following mistakes should be avoided during calculations 1. Wrong transfer of data (WT) This is as a result of rounding off at different stages. For instance Sample question 57 A sample of a limestone reacts with 22.5cm3 of 0.2M HCl. Calculate a. The number of moles of the acid which reacted. Moles = Molarity x Volume, = 0.2 x 22.5 = 0.0045moles 1000 b. The mass of CaCO3 in the limestone. CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) Mole ratio 1 : 2 0.005 0.0045 2 =0.0025 moles of CaCO3 Mass = 0.0025 x 100 ( Molar mass of CaCO3) = 0.25g. From the calculation above the student will fail due to WT - wrong transfer of data; 0.0045 is rounded off to 0.005 moles. When divided by 2 we get 0.0025 moles instead of 0.00225 moles. 95

Tactics of Chemistry 2. Strange Values ( SV) This is introducing a value whose origin is not known. Sample question 58 A current of 3A passed for 10 minutes 20 seconds to deposit copper from copper (II) sulphate solution. Calculate the amount of copper deposited. ( 1F= 96,500, Cu = 63.5) Quantity of electricity passed = Current x time = 3 x {(10 x60) +20} = 1,860 Coulombs Cu2+(aq) + 2e Cu(s) To deposit one mole of Cu, two moles of electrons are required. The charge required = 2 x 96, 500 = 193,000C Mass of copper deposited = 1,860 x 63 193,000 = 0.6072g. In this question, the value 63 is a strange value. Where is its origin?. 3. Whole number mole ratio. The following mole ratio should always be a whole number. a. The Water of crystallization. Sample question 59 When a sample of hydrated sodium carbonate salt was heated in open dish, the mass reduced by 62.94%. Write the empirical formula of the hydrated sodium carbonate. (Na=23.0, C=12.0, O=16.0, H=1.0) Answer. The percentage of water = 62.94 that of Na2CO3 = 100 - 62.94 = 37.06% 96

Learning chemistry made easier NaCO3 H2O The amount of 37.06 62.94 water is expressed 106 18 to the nearest whole number, 0.3496 3.4967 as 10 instead of 0.3496 0.3496 10.002 1 10.002 1 : 10 See other salts with whole number mole ratio in hydrated form • (NH4)2SO4.FeSO4.6H2O - Ammonium iron (II) sulphate • MgSO4.7H2O - Magnesium sulphate salt • Al2(SO4)3.16H20 - Aluminium sulphate • CuSO4.5H2O - Copper (II) sulphate • Na2S2O3.5H2O - Sodium thiosulphate • Fe(NO3)2.9H2O- Iron (II) nitrate • (COOH)2.2H2O - Ethan-1,2-dioic acid/ oxalic acid. b. Number of molecules in an organic compound. c. Vmaoluleeroaftinoionfathcearabcoidx:ybliacsaeciisdcia.elc(uClOatOeHd)tnh. eIfvtahleufeinoafln must be a whole number. Sample question 60 The compound below has a molecular mass of 2490g. Calculate the value of n. HH C-C HHn One molecule has 28g Number of molecules = 2490 = 88.92, = 89 molecules. 28 97

Tactics of Chemistry Parallel work is not marked unless logical formulae are used to achieve the same objective and answer. The Tactic An examiner is aware of the students who are uncertain or who don`t know the way to work out a question, will try to confuse the examiner by using parallel working to earn marks by benefit of doubt Sample question 61 10g of Chlorine -39 decays to 1.25g in 165 minutes. What is its half life? Method 2 N= Method 1 This student is N0= (1/2)T/t N0 (1/2)n torn between the N correct formula 1 & 2. In spite of the two working (1.25/10) = (1/2)n arriving to the (10/1.25) = (1/2)165/t (1/8) = (1/2)n same answer, (8/1) = (1/2)165/t (1/2)3 = (1/2)n there is serious arithmeticerrors in method 2. (2/1)3 = (1/2)165/t n = T/t Learn to use the n = T/t correct formula, you cannot mix t = T/n = 165/3=55minutes up the examiner. t = T/n = 165/3=55minutes Cancelled work is not marked regardless whether the answer of the cancelled work is correct. Sample question 62 15g of MgCl2 salt saturates 25cm3 of H2O at 250c. Find its solubility Method 1 Solubility is expressed Method 2 per 100g of water. 15g saturates 25g of water 15g saturates 25g of water The solubility of MgCl2 is This students cancels To saturate 100g of water the correct answer. Method 1 is wrong. = /15 Method 2 is correct = /15 x 100 25 25 but cancelled. No = 0.60g/1g of water marks scored. = 60g/100g of water 98

Learning chemistry made easier Tactical Practise L 1. 100cm3 of 0.05M sulphuric acid were placed in a flask and a small quantity of anhydrous sodium carbonate added. The mixture was boiled to expel all the carbon (IV) oxide. 25cm3 of the resulting solution required 18cm3 of 0.1M sodium hydroxide to neutralise it. Calculate the mass of sodium carbonate added. (Na=23.0, O=16.0, C=12.0) (3 marks) 2. An organic compound had the following composition 37.21% carbon, 7.75% hydrogen and the rest chlorine. Determine the molecular formula of the compound, given that the molecular mass of the compound is 65. (H=12.0; H=1.0; Cl=35.5) (3 marks) 3. A solution was made by dissolving 8.2g of calcium nitrate to give 2 litres of solution. Determine the concentration of nitrate ions in the moles per litre. (3 marks) (Ca=40.0; N=14.0; O= 16.0) (3 marks) 4. When a stream of hydrogen gas was passed over 5.76g of heated Copper II oxide, 0.72g of water was produced. Calculate the simplest formula of the copper oxide. (3marks) 5. A beaker contained 75.0cm3 of aqueous copper (II) sulphate at 23.70C. When scrap iron metal was added to the solution, the temperature rose to 29.30C. Given that the mass of copper deposited was 5.83g, calculate the molar enthalpy change in kJmol-1. ( specific heat capacity of solution=4.2Jg-1K-1, density of solution=1gcm-3, Cu=63.5) (3 marks) 6. Analysis of a compound showed that it had the following composition: 69.42% carbon, 4.13% hydrogen and the rest is oxygen. a. Determine the empirical formula of the compound. (C=12.0, H=1.0, O=16.0) (2 marks) b. If the mass of one mole of the compound is 242, determine its molecular formula. ( 1 mark) 7. 22.2cm³ of sodium hydroxide solution, containing 4.0g per litre of sodium hydroxide were required for complete neutralization of 0.1g of a dibasic acid. Calculate the relative formula mass of the dibasic acid (Na = 23.0, O = 16.0, H = 1.0). (3marks) 8. X grams of a radioactive isotope take 100 days to decay to 20g. If half life of the element is 25 days, calculate the initial mass X of the radioisotope. (2marks) 9. When a current of 6.42 A was passed through an electrolyte containing Y2+ ions for 10 minutes, 2.74 of Y were deposited i. Calculate the quantity of electricity passed in the experiment. (1 mark) ii. Determine the relative atomic mass of Y (1Faraday = 96,500 coulombs) (3 marks) 99

Tactics of Chemistry 10. When nitrate solution of a certain metal X was electrolysed, 1.174g of metal X was deposited by a current of 4 Amperes flowing for 16 minutes. Determine the formula of the metal nitrate. (1F= 96,500, R.A.M of X= 59) (3 marks) 11. An element Q consists of 3 isotopes of mass 28, 29, 30 and percentage abundance of 92.2, 4.7, 3.1 respectively. Determine the relative atomic mass of the element? (3 marks) 12. Natural Gallium consists of two isotopes, with atomic masses 69 and 71 in the atomic ratio of 3: 2 respectively. Calculate the relative atomic mass of Gallium. (2 marks) 13. 10 molecules of an unknown gas have a mass of 1.0667 × 10-21 g. Determine the relative molecular mass of the gas. (L=6.0×1023) (2 marks) 14. 10g of Chlorine -39 decays to 1.25g in 165 minutes. What is its half life? (2marks) 15. Calculate the number of Al3+ ions released when 30cm3 of 0.1M of Aluminium Sulphate is dissolved in water. (L = 6.0 x 1023). (3 marks) 16. A factory produces 63.6 tones of aonfhtyodnreosusoNf as2oCdOiu3 mon a certain day by Solvay process. Calculate the number chloride used on this particular day. Assume the plant is working at 100% efficiency. (C = 12, H = 1, Cl = 35.5, Ca = 40, Na = 23) (3 marks) 17. 5 ml of 1M potassium iodide solution completely reacted with 2.5ml of 1M lead (II) nitrate solution. Use this information to calculate the number of moles of iodide ions that reacted with one mole of lead (II) ion. ( 3 marks) 18. When excess dilute hydrochloric acid was added to sodium sulphite, 960cm3 of sulphuric (IV) Oxide gas was produced. Calculate the mass of sodium sulphate that was used. (Molar gas volume = 24000cm3 and S = 32 Na = 23, O = 16). (3 marks) 19. 3.31g of lead (II) nitrate was strongly heated i. Write the equation for the decomposition that occurred. (1mark) ii. Calculate the mass of the residue formed (2marks) (Pb = 207, N = 14, O = 16) 20. A bottle containing concentrated ammonia solution has a label with the following information. Density = 0.880g/cm3 Percentage purity by mass = 32%. Calculate the concentration of ammonia solution in moles per litre (N = 14, H = 1) (3marks) 100

Learning chemistry made easier 5. GRAPHS Graphs are very vital in chemistry ; graphical knowledge, techniques or skills are required. Graphs are highly recommended because they are very analytical and able to present data in a simple and a logical or rather descriptive way. A graph can predict a decrease, increase etc or provide a wide range of data from a small range. The following are the checking points for graphs • Labelling of axis (L) • Correct scale (S) • Plotted points (P) • Curve/line (C) Title Title is not that important but there is no harm indicating it. You may state your title depending with what you are plotting. For instance solubility curve for potassium nitrate, the graph of pressure against volume of a gas etc Plotting Plotting is tracing the points from the table of values to the graph paper. Plotting can be done using • a cross (X) • a visible dot. ( ) Lets look at the two plotting methods. Using dots to plot the points on the graph. 101

Tactics of Chemistry Using crosses to plot the points on the graph. Avoid tracing using invisible dots or circles. Invisible dots might be covered by the line or the curve and therefore appear as a mere line or curve. Chemistry data uses small portions and therefore the range is small a reason why plotting using circles is not recommended. Plotting usually carries two marks or more. The Tactic A line or a curve on a graph with no visible plots earns ZERO marks. The two wrong methods which should not be used in chemistry are: Using invisible dots to Using a circle to plot plot the points on the the points on the graph. graph. 102

Learning chemistry made easier If the graph starts from the origin, you have to plot at the origin too. Plot the origin if the graph passes through origin. Avoid inversion of the graph; it will not be marked or else you are penalized. Inversion is when Y- axis is plotted as X-axis and the vice versa, for instance, If you are required to plot a graph of volume aingathinesgt r1a/ptimhe then volume should be on y - axis and time on x–axis but below it has been inverted. The Tactic The word against is very general and does not specify which quantity is plotted on which axis. Most of the graphs are usually specified which quantity on which axis. If not specified; all dependent quantities are plotted on x-axis while the independent quantities on y-axis. When plotting time versus concentration, the time is dependent on concentration. (sec-1) An inverted graph 1/ Volume (cm3) Time 103

Tactics of Chemistry When you do the graph, the plotting predicts what is expected, whether a line or a curve. An alternative method is using your skills on laws which give the proportions to predict the graph. If a graph has 5 points to be plotted, the marking is done as follows. • 4 or 5 points correctly plotted, you earn full marks. • 3 points plotted correctly you earn half of the allocated marks. • 2 correct points earn half a mark while 1 point earns no mark. • Remember your points must be rounded off to 3 decimal places incase the values plotted are not whole numbers. Curve/Line The graph should be a single thin smooth line passing through the plotted points which only align to either as a straight line or as a curve. Smooth curve as predicted by the plotting For a correct graph curve/line should pass through almost all the plotted points. Wrong data results to a staggered plotting. Avoid drawing your line/curve on such points because it does not show the proportionality of the quantities. Look at this staggered graph A staggered plotting of a graph. Some points might be correct and others not. Do not draw rather draw a straight line graph. 104

Learning chemistry made easier Labeling Labelling should be done along the axis, to show the quantity being represented. The quantity and the units are indicated in brackets after labelling, as shown in the graph below. Volume (cm3) Time (s) Proper units must be indicated on the axes. If wrong units are used you lose marks. If you label your axis correctly, but fail to indicate units you are penalized. This is the same as giving wrong units. Scale Scale determines the size of your graph. The following are preferred conditions for your scale. • Should be simple, clear and consistent on both axes. 1p/a2goef the • Must poyrcoocuvuripdsyceadalteislbeeraetstcteor3m/a4nmodefntyhdoeeudgrr, appuahgsielna, grthgeo3r/u.4ghofevthene will page make Incase your graph is too large and might not fit on the page provided, the break of axis is acceptable. Incase you chose to break the axis ensure that their is no question required to be answered using the broken portion. Any axis can only be broken once and from the point of origin. 105

Tactics of Chemistry Solubility in g/100g of water Broken axis Temperature (0C) Other sketch graphs which you must understand how they work are the graphs involving reaction rate like in the example below. Sample question 63 Two - 2cm long magnesium ribbons were reacted with 50cm3 of 2M and 1M hydrochloric acid in a separate beakers and timing done simultaneously. Sketch a graph to show the liberation of gas against time up to the end of the experiment. 2 M HCl 1 M HCl Volume of H2 (cm3) Time (s) 106

Pressure (Pa) Learning chemistry made easier Volume (cm3)In this questions you are require to sketch a graph. There is no need to show the plotting because there is no table of values. In such as graph the volume of hydrogen gas evolved is the same but end points of the reactions are different due to different concentrations of acid. Other questions which requires you to sketch are gas law questions. Sample question 64 Sketch a graph showing the relation ship between pressure and volume of a fixed mass of a gas. Volume (cm3) Sample question 65 Sketch a graph showing the relationship between temperature and the volume of a fixed mass of a gas in the axis below. 0 Temperature (0C) Questions on reaction rates are very common; where the production of volume is measured at different time intervals. You must ensure that the graph joins with each other if the same mass of the solid was 107

Tactics of Chemistry used. In such a graph, the gas evolved is the same but time for the completion is different due to different; • Concentrations • Temperature • Surface area • Catalyst • Pressure The two curves must join with each other at the completion of the reaction. This is because they end up giving the same volume of the gas. The only difference is the rate at which the gas is produced in each case. To calculate for the rate of reaction at a point, draw the tangent at that point and determine the gradient of the line at that point ie Rate of reaction = Change in y = /Δy Change in x Δx Trends of a data The data collected during an experiment have a particular trend. This depends on the nature of the reactants. When reacting an acid and a base it is expected that as the reaction progresses the temperature rises gradually to the highest point then fall after the end point is attained. This behaviour of the temperature is it`s trend for that reaction. When required to fill a table during an experiment, do note that a certain trend is expected. This trend is usually awarded marks. If your data does not have a particular trend in line or which agrees with a fact, then there may be a problem with that data of the experiment. If the reactants are mixed and there is an expected temperature change as time progress then the trend is either • There is a gradual temperature rise the it fall after some times. or: • There is a gradual temperature fall followed by a rise after sometime. 108

Learning chemistry made easier For questions which have the table of values is important to draw a good graph on the graph papers provided by the examiner. Lets look at a well done question on graphs. Sample question 66 In an experiment to investigate the rate of reaction between equal masses of powdered and a lump of zinc carbonate with excess 2M hydrochloric acid, the following data was recorded during experiment. Time (s) 0 5 10 15 20 25 30 0.0 10.0 19.0 22.5 24.0 24.0 24.0 Volume of CO2 using powdered ZnCO3 0.0 4.0 9.0 13.5 18.0 22.0 24.0 Volume of CO2 using lumps of ZnCO3 Plot a graph of volumes (y-axis) against time. Volume of CO2 (cm3) 25 Powdered 20 Lump 15 10 5 0 5 1T0ime 15 20 25 30 35 (s) 109

Tactics of Chemistry 1. Label the two curves When labeling the curve/line avoid showing another line from a graph to the word. This may be regarded as part of the graph. On the graph below the line showing powdered and lump graphs can be considered as part of the graph. Avoid that. Volume of CO2 (cm3) 25 Powdered Lump 20 15 10 5 0 5 10 15 20 25 30 35 Time (s) 2. Explain the difference in shapes of the curves The powdered zinc carbonate have a larger surface area in contact with the acid and therefore higher rate of reaction compared to lumps of zinc carbonate of the same mass. 3. What would happen if the powdered solid was reacted with 4M hydrochloric acid. The rate of reaction will be faster due to higher rate of successful collision. 110

Learning chemistry made easier From the graph, determine a. the volume for CO2 of powdered zinc carbonate at: i. 7 seconds = 14.5 cm3 ii. 13 seconds = 21.5 cm3 b. Calculate the rate of reaction of the lump at 20 seconds Rate of reaction = Change in volume Change in time = 23.0 - 15.0 = 8.0 = 0.889cm3s-1 25.0 - 16.0 9.0 Tactical Practise M 1. In an experiment on rates of reaction, potassium carbonate was reacted with dilute sulphuric (VI) acid. a. What would be the effect of an increase in the concentration of the acid on the rate of reaction? ( 1 mark) b. Explain why the rate of reaction is found to increase with temperature. (2 marks) 2. The results below awtesrteanodbatardinteedmwpheerantu2r.5egaonfdMpCreOs3swureer.e reacted with excess 2M nitric (V) acid Time (sec) 0 2 4 6 8 1 12 14 16 18 Volume of gas (cm3) 0 165 270 350 410 445 465 480 480 480 a. i. Draw the graph of volume of gas against time (x–axis). (3marks) ii. From the graph, find the time the reaction stops. (1mark) iii. What is the maximum volume of the gas evolved? (½ mark) iv. Determine the rate of the reaction at time t = 4 second. (1mark) b. i. Write down the equation for the reaction (1mark) ii. DHeetnecrem, fiinnedtthheemreollaatrivme aastosmoficMmCOas3s. (o2fmMa.r (k2sm) (aCr=k1s2) , O=16) iii. 3. The sketch below represents a graph obtained when zinc granules were reacted with excess 0.2M Sulphuric (VI) acid in the presence of a catalyst in a conical flask placed on an electronic balance. 111

Tactics of Chemistry Loss of mass (g) Time(s) i. Write an equation for the reaction that took place. (1mark) ii. Explain why there is loss in mass. (1mark) iii. Name the catalyst used. (1mark) iv. Sketch, on the same axes, the curves obtained when: I. the same mass of zinc powder was used under the same conditions. (1mark) II. no catalyst was used. (1mark) 4. Sodium thiosulphate solution reacts with dilute hydrochloric acid according to the equation. S O -3 + 2H+(aq) H2O(s) + SO2(g) + S(s) 2 2 (aq) In an experiment to study how rate of reaction varies with concentration, 10cm3 of 0.4M sodium thiosulphate was mixed with 10cm3 of 2M Hydrochloric Acid in a flask. The flask was placed in a white paper marked with a cross X. The time taken for the cross (x) to be invisible when viewed from the above was noted and recorded in the table below. Experiment Volume of 0.4M Moles of Volume of Volume of Time in thiosulphate thiosulphate used water 2MHCl seconds 1 10 0 10 16 10 23 2 7.5 2.5 10 32 10 3 5.0 5.0 72 4 2.5 7.5 a. i. Plot a graph of volume of thiosulphate (vertical axis) against time taken for cross to become invisible. (4 marks) ii. From the graph determine how long it would take for the cross to become invisible if the experiment was done. 112

Learning chemistry made easier I. Using 6cm3 of 0.4M thiosulphate (1 mark) II. Using 60cm3 of 0.2M thiosulphate (1 mark) b. i. Using values from the experiment Calculate I. Moles of thiosulphate used (1 mark) II. Moles of hydrochloric acid used. (1 mark) ii. Explain which of the two reactants in the experiment I in b (i) above controlled the rate of reaction. (2 marks) c. Give two precautions which should be taken in this experiment.(2 marks) 5. The table below describes three experiments on the reaction of excess dilute sulphuric (VI) acid and 0.5g of zinc, done under different conditions. In each case volume of gas produced was recorded at different time intervals. Experiment Physical appearance of Dilute sulphuric (VI) Zinc acid I Powder 0.8 M II Powder 1.0 M III Granules 0.8 M i. On the axes below, sketch and label the three curves that would be obtained from these experiments Volume of Hydrogen gas (cm3) Time (min) ii. Explain your answer for the curves sketched above. 113

Tactics of Chemistry 6. SCIENTIFIC DATA REPRESENTATION Some data requires to be treated in a particular way depending on the laws and the principles the data has to fulfil. The given unit helps you decide the type of treatment to be done on the raw data. Either to use the data in the given units or change it to other units to fit the laws and principles governing scientific facts. Such data requiring treatment includes: • Volume - In most questions, volume is always as a ratio and conversion may not be necessary. In gas law questions for instance, it is not important to change the volume unless the volumes given are in different units or the is answer required in a different unit of volume as specified. • Temperature - When dealing with the gas laws, temperature must be changed from 0C ( degree Celsius) to K (Kelvin) by adding 273 to 0C. This is because ideal gas laws are only applicable in absolute temperature scale. Example 250C + 273 = 298K In enthalpy changing temperature is not necessary since the difference in both Celsius and Kelvin is the same. For instance, if the temperature of a reaction changes from 200C to 300C, the difference is 30-20 = 100C. In absolute scale (273 + 30) - (273 + 20) = 10K. Sample question 67 20cm3 of a gas is heated from 320C to 800C at constant pressure of 760mmHg. Calculate the new volume Using Charles` law V1 = 20cm3, T1 = 32 +273 = 305K, T2 = 80+273=353K V1 V2 V2 20 x 353 = 23.15cm3 T1 T2 305 114

Learning chemistry made easier • Molarity - Always this concentration in to moles per litre (M). Most of times volume is usually given in cm3, it is important to change in into litres. Sample question 68 3.0g of sodium hydroxide pellets were dissolved in 50cm3 of water. Calculate the molarity of the solution. Molarity = Moles per litre = 0.075 moles Volume Moles = Mass Molarity = 0.075 x 1000 Molar mass 50 = 3.0 = 1.5moles/litre or 1.5M 40 • Concentration - Can be used in grams per litre or moles per litre (molarity), depending on the mode of the question or as specified in the question. Sample question 69 3.0g of sodium hydroxide pellets were dissolved in 50cm3 of water. Calculate the concentration in grams/litre of the solution. 3g are contained in 50cm3 ? g are contained in 1000cm3 Concentration = 3 x 1000 = 60g/litre. 50 • Electric Charge -This is the electric current flowing through a conductor in a given time. It is given in coulombs, C. Electric charge, Q is a product of electric current ( I) and time (t). Both current and time must be expressed in their SI units; ie seconds, (s) for time and Amperes (A) for electric current. Mathematically charge, Q it is expressed as Q = It , where I is current in Amperes, t is time in seconds. While Q is the quantity of electricity in Coulombs. 115

Tactics of Chemistry Sample question 70 A current of 5A passes through brine for 3minutes and 20seconds. Calculate the electric charge passed. Q = It = 5 x {(3 x 60) + 20} = 1,000Coulombs • Enthalpy change - when calculating the enthalpy of a reaction, it is important to check the units of specific heat capacity. For instance, if the specific heat capacity of water is 4.2kJKg-1K-1, 4.2Jg-1K-1 or 4,200JKg-1K-1 then the given mass must therefore change to coincide with those of specific heat capacity. Sample question 71 6Tgheomf KixNtOur3eswolaids was added to 120cm3 of water in a plastic beaker. stirred gently and the following results obtained. Initial temperature = 21.50C Final temperature = 17.0 0C a. Calculate the enthalpy change for the reaction (Density of water=1g/cm3, and Specific heat capacity of water c= 4.2jg-1K-1) Heat = Mass x Specific heat capacity x temperature change Note that c of water is = (120 x 1)g x 4.2 x (21.5-17) already given in 4.2Jg-1K-1 = 120 x 4.2 x 4.5 and you don`t need to convert the mass in Kg. Ensure you use =2,268J the units as specified. b. Calculate the molar enthalpy change for the dissolution of potassium nitrate. (K=39, N= 14, O =16) = 6g of KNO3 produces 2,268J of heat energy = = 0.05941 moles of 2,268J of energy /6 KNO3 produces 101 1 mole of KNO3 produces = /2,268 x 1 0.05941 = 38,175.39Jmol-1 or 38.175kJmol-1 116

Learning chemistry made easier SECTION B PAPER 233/3 PRACTICAL PAPER

Tactics of Chemistry CHEMISTRY PAPER THREE (PRACTICAL) This is the chemistry practical paper and has a total marks of 40. A practical paper is very important because the examiner ensures the paper contributes a higher percentage. Remember, science is done by; • Observation • Experimentation A practical paper can emanate from any area of chemistry. This makes the theory papers very important for capturing of the concepts. The most frequent examinable areas are: • Titration/Volumetric analysis/Quantitative analysis • Qualitative analysis/test of ions • Physical Chemistry • Enthalpy change • Rates of reaction. • Solubility of salts 1. TITRATION Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurement plays a key role in titration, it is also known as volumetric analysis. A reagent called titrant or titrator is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrant to determine concentration. Titration is of two types • Acid - base titration - These are titration which require an indicator to note the end point. They include normal or the so called neutralisation and back titration. 118

Learning chemistry made easier • Redox titration- these are titration which do not require an indicator to note the end point. Solution changes colour at end point. The volume of titrant reacted is called titre. The trick for titration is • Proper rinsing of the apparatus. This is important because contaminated apparatus will lead to errors in titration. • Noting a constant end point - This happens when a colour of the solution in the conical flask appears permanently or disappears permanently regardless of how fade the colour is. • Colour noting is a problem to many students, after a faint permanent colour appear students continue titrating until the colour becomes completely dense. This is wrong, because the volume of the titre will have increased. For instance when titrating NaOH with HCl using phenolphthalein indicator in the acid, the following colours appear. Titration in progress The reactants Temporary Fade permanent Dense permanent are ready for colour change colour change is colour change is titration. is observed. observed. observed. The exact point to take your reading should be when the first permanent colour change is observed, regardless of how fade the solution is. Knowledge on what titration is and the way it is carried out is more important for it helps you understand the flow of questions. 119

Tactics of Chemistry Terms Used in Titration. 1. Standard Solution - This is the solution whose concentration is known. 2. Titrant - This is the solution in the burette which titrates the standard solution. 3. Titre - This is the volume of the titrant which takes reaction to completion (end point). 4. Indicator - This is a substance that marks/indicates the end point of a reaction. 5. End point -Also known as equivalence point. This is a point where the reaction has reached completion. It is indicated by a permanent colour change. 6. Analyte - This is the solution in the reaction vessel and whose volume is known. Analyte is usually the volume measured using a pipette and put in a conical flask for titration. Apparatus for titration. They include: burette, pipette, conical flasks, volumetric flask, funnel, complete retort stand, pipette filler, dropper, white tile etc Tactics when using the Apparatus All the apparatus MUST be rinsed with distilled water before using them whether sharing or not. The droppers provided in the solutions should not be moved or alternated to avoid contamination. 1. Burette - It contains the titrant. Use the titrant only in the calibrated part of the burette. Let the solution run drop wise at your own manageable pace in to the standard solution. 2. Pipette - Fill the solution up to the mark. When transferring the solution DON`T blow the remaining solution at the end in stead let it run on its own. It may results to aliquot. 120

Learning chemistry made easier 3. Conical Flask - It should be whirled not swirled. Swirling causes the solution to splash on the wall of the conical flask which will increase the errors in volume. 4. Volumetric flask - Always fill the solution or the given liquid up to the mark. 5. Funnel - This assist in filling the burette. Ensure you remove the funnel immediately you are through filling the burette. The funnel might have drops of the solution which change your reading of the burette. 6. Complete retort stand - Use clamp well, don`t tighten it so much when the burette is held. The burette will break. 7. Pipette filler - This is a plastic rubber which usually replaces the use of mouth. Care should be taken when using a pipette filler so that the solution does not get in to the filler. Incase you are not sure of how it is used ask the teacher or the laboratory technician. Using distilled water, do practice with pipette filler to perfect with how operates. See the fillers below. This is how pipette filler is fixed 8. Dropper - Use the corrects number of drops as specified. The dropper should be held vertically so that the indicator drops in to the solution without getting to the sides of the container. Avoid mixing droppers. 121

Tactics of Chemistry In titration, the following are the areas under test or the marking areas. 1. Filling tables 2. Use of decimals 3. Accuracy & precision 4. Principles of averaging 5. Final answer 6. Calculations i. Filling of Tables Always ensure you fill the required titration with concordant titre value which do not vary by more than + 0.2cm3. Using this precision of + 0.2cm3 a titration table can be filled in two ways. Both ways are correct but the Table I I II III 10.10 20.20 30.30 Titration 0.00 10.10 20.20 Final Burette reading (cm3) 10.10 10.10 10.10 Initial Burette reading (cm3) Volume of titre (cm3) On table I, the difference among the 1st, 2nd and 3rd titres maintains the standard within + 0.2cm3. The difference between the highest titre and the lowest also maintains a precision of + 0.2cm3. The three titre values must be averaged in this case. Table II Titration I II III Final Burette reading (cm3) 10.60 21.00 31.20 Initial Burette reading (cm3) 0.00 10.60 21.00 Volume of titre (cm3) 10.60 10.40 10.20 On table II, a precision of + 0.2cm3 is maintained in 1st, 2nd and 2nd, 3rd. The range between the highest 10.60 and the lowest 10.20 is 0.4cm3. The student can average either the first two tires or the last two titres in this case. 122

Learning chemistry made easier The Tactic In both tables a precision of + 0.2cm3 is maintained. This is accepted. A lot of time, students do fake the other titre as same with the first one without performing the required titration. It is always very significant to perform all the expected titrations to minimize the errors. Don`t judge to conclude the titre with the first titration. It might be wrong. The table marks are awarded as follows • Complete table with all three titrations ( 1 mark) • Incomplete table with two titrations ( 01/m2 marakr)k) • Incomplete table with one titration ( The penalties set on filling of the table are; • Wrong arithmetic on subtractions • Unrealistic values either too high like 75cm3 or too low i.e. below 1cm3 without explanations. • Burette reading above 50cm3 without explanation. • An overturned or an inverted tables score zero marks. An inverted table is a table where the final readings have been interchanged with the initial burette reading. • Inconsistency of decimals like using 15.0, 15 and 15.00 together... An example of a poorly filled table. Titration I II III Final Burette reading (cm3) 0 23.5 45.5 Initial Burette reading (cm3) 10.0 10.00 23.5 Volume of titre (cm3) 10 13.5 15 Mistakes - Wrong precision, Inconsistency of decimal points, Wrong arithmetic and Inversion, In case of wrong arithmetic/subtractions in the table you are penalized accordingly. 123

Tactics of Chemistry ii. Use of decimals All your reading from the burette should be accurate and consistence to one or two decimal places. If two decimal places are used, then the second decimal place must be 0 or 5 e.g 30.10cm3, 35.55cm3 and not 30.01cm3, 35.53cm3 since the accuracy of the burette does not allow such measurements. This is an example of a well filled table with proper use of decimals. Titration I II III Final Burette reading (cm3) 10.60 21.00 31.20 Initial Burette reading (cm3) 0.00 10.60 21.00 Volume of titre (cm3) 10.60 10.40 10.20 If any of the values in rows are inconsistent as per the decimal places penalty is full, zero mark. 0 mark is awarded for decimal places. Ensure that your averaged titre values have a precision of ±0.2, For full credit of marks. iii. Accuracy & Precision 1 mark is awarded for value within + 0.1 of the school value. 1/2 mark is awarded for any values within ±0.2 to the school value. If no titre value is within ±0.2 of the of school value (SV), 0 mark is credited. iv. Principles of averaging • Don’t attempt to average values which are not within ±0.2cm3. Be aware that when you are doing your KCSE practical, your teacher does the same practical in a different room to obtain school value. These are the marking principles. • If three concordant values are averaged - 1 mark is awarded. = 15.90 + 15.80 + 15.80 3 = 15.83cm3 = 47.50 124 3

Learning chemistry made easier • If three titration are done and only 2 are concordant within a range of ±0.2cm3, then only the two are averaged full marks of (1mark) is awarded. • If 3 done titrations are concordant and have a precision ±0.2cm3 , but only two are averaged (0 mark) • If 3 titration are done and are inconsistent and yet averaged ( 0 mark) • If no working is shown you are penalized fully to get 0 mark. • Rounding off the average volume to 2 decimal places is accepted. 1/2 mark is penalized for rounding off to 1 decimal place unless exact division is realized. v. Final answer A student`s average titre is compared with the school value, If within ±0.1 to the S.V ….. Full mark credit If within ±0.2 to the SV …….1/2 mark credit. If there are other values that the student would have averaged and still be within the required range, then the marks are credited accordingly. If a student gives two sets of averages, the principle of averaging marks is penalized fully. N/B The student should only average once. Exaple: Student`s average titre 25.6cm3 25.4cm3 25.2cm3 You cannot average the three titre because they are not within ± 0.2cm3. Though the 1st and 2nd are within the range as well as 2nd and 3rd, but 1st and 3rd are out of range. If you opt to choose two, Student`s average 1. 25.2 + 25.4 = 25.3cm3 2 Student`s average 2. 25.4 + 25.6 = 25.5 cm3 2 125

Tactics of Chemistry If the school value is 25.6 cm3, you get full credit since one of the averaged answer 25.5 is within ±0.1 of the school value. If the answer is within + 0.2 of the school value then 1/2 mark is credited. Student who use style of table II are in a greater possibility to score better than those who use style of table I unless it is exact and correct.. vi. Titration calculations This tests the general practical & theoretical understanding of the mole concept. You are encouraged to work from principles. The applied principles are: 1. Dilution Principle This is where a more concentrated solution is diluted with distilled water to get a lower concentrated solution. M1V1 = M2V2 Where MVVM1221----VVMMooollouullammarreeiittyyooffooahffiddhgiiihlgluuehtrteeecrddocssnooocnlleuucnetttiinorotannrt.aetdedsosloultuiotinon Sample question 72a. You are provided with 2.5M sulphuric (VI) acid solution A. Pipette 25cm3 of solution A in a 250ml volumetric flask. Add distilled water to the mark. Shake well and label this solution D. Calculate the molarity of solution D. m1v1 = m2v 2 m2 = 2.5 x 25 = 0.25M 250 2. Acid - Base mole ratio This applies to questions which involve acid-base titration where Molarity of acid x Volume of acid = Acid:base mole ratio Molarity of base x Volume of base 126

Learning chemistry made easier Ma x Va = Acid:base Mb x Vb mole ratio Sample question 72b. 18cm3 of solution D reacted completely with 23cm3 of sodium hydroxide solution. Calculate the molarity of sodium hydroxide. Answer. Ma = 0.25M, Mb = ?, Va = 18cm3 and Vb = 23cm3 H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H20(l) Mole ratio 1 : 2 Using the acid : base mole ratio equation 0.25 x 18 1 Mb = 0.391M Mb x 23 2 3. Moles formula This is a very common formula Moles = Mass given Molar mass Sample question 73a. Solution K was prepared by dissolving 12.5g of potassium nitrate in 250cm3 of water. How many moles of potassium nitrate are contained in 250ml of water. Moles = Mass given Moles = 12.5 = 0.01238moles Molar mass 101 4. Molarity formula 127

Tactics of Chemistry Molarity = Moles Volume Sample question 73b. Calculate the molarity of the solution K. Molarity = Moles Molarity = 0.01238 x 1000 Volume 250 = 0.0495M Common mistakes in calculations arise from students inability to comprehend the flow of question using the data provided or obtained by the student. During calculations, the following standards are checked: • Concentration ( molarity) should be to at least 3 decimal places unless exact. • Moles should be at least in 4 decimal places 0.045648 = 0.0456 and not 0.046 unless exact. • Where units are expected they should be present and correct unless they are implied in the stem of the question. For instance = 9.23 Average volume in cm3 = 9.2 + 9.3 + 9.2 3 • Note that working beyond the expected answer is penalized fully and zero mark is awarded. • Avoid strange figures, they may be termed or regarded as a collusion or conspiracy either between student-student (S&S) or Teacher–Student (T&S). This may lead to cancellation of your paper due to cheating. Sample question 74 Consider the following table filled by a student. 128

Learning chemistry made easier Titration I II III Final Burette reading (cm3) 15.1 30.10 45.1 Initial Burette reading (cm3) 0.00 15.1 30.10 Volume of titre (cm3) 15.1 15 15.00 Suppose the school value (S.V) is 15.0cm3 Calculate the average titre Average Titre = 15.1 + 15.0 = 15.05 cm3 2 When Marking • The student gets full marks for performing the three titration. • The student loses full mark due to inconsistency of decimal places. The reading of the 2nd titration has no decimal places unlike others that have one decimal place. • Student earns 1 mark for accuracy. • On averaging, the student loses 1 mark for only taking two values and excluding the third value which is within 0.2cm3 range with the others. • Even if the average titre is within the range of the school value (S.V), the student loses because he did not average the three titres within ±0.2cm3. N/B. Inconsistency in decimal places is checked in initial and final burette reading only. Inconsistency in initial burette reading as 0, .0 or 0.00 is not penalized. Let us go through the following redox titration step by step. Sample question 75 You are provided with the following. • A solution S, made by dissolving 4.9g of FeSO4(NH4)2SO4.6H2O in 250cm3 of solution. • Solution R, a solution of Potassium Manganate (VII) containing 0.002 moles in 100cm3 of solution. 129

Tactics of Chemistry You are required to determine the mole ratio of R to S and write a balanced ionic equation for the reaction that occurs. Procedure • Fill the burette with solution R. Pipette 25.0cm3 of solution S in to a clean 250ml conical flask. Add 5cm3 of 2M sulphuric (VI) acid in to the conical flask containing solution S and titrate with solution R from burette till the permanent pink colour appears. Repeat the procedure two more times and record your results in the table below. Note that this value may vary from one school to the other, depending on the chemicals and the preparation. The school for this titration is value is 12.50cm3. Titration I II III Final Burette reading (cm3) 12.60 25.00 37.50 (4 marks) Initial Burette reading (cm3) 0.00 12.60 25.00 Volume of titre (cm3) 12.60 12.40 12.50 The marking criteria is done as follows • Done all titration - 1 mark • Done two titrations -01m/2amrkark • Done one titration - • Consistency of decimal -1mark • Accuracy - 1 mark • Principle of averaging - 1 mark • Final answer - 1 mark The student satisfies the marking criteria for the 4 marks. Calculate i. The average volume of solution R Note = 12.60 + 12.40 + 12.50 • The averaged titres have 3 range within + 0.2cm3 = 37.50 3 • The average titre is within = 12.50cm3 + 0.1cm3 of the school value; SV 130

Learning chemistry made easier ii. Concentration of R in moles per litre = 0.002 moles are contained in 100cm3. Molarity = Moles Volume = 0.002 x 1000 100 = 0.02M is the number of moles of R that reacted Volume used = 12.50cm3 and Molarity = 0.02M Moles = Molarity x Volume Moles = 0.02 x 12.50 1000 = 0.00025 moles iii. Concentration of S in moles per litre Mass used = 4.9g Molar mass of FeSO4 (NH4)2SO4.6H2O = 392g Moles = 4.9 Alternatively 392 4.9g are in 250cm3 = 0.0125 moles 1000cm3 will contain If 0.0125 moles are in 250cm3 =4.9 x 1000 ? moles are in 1000cm3 250 = 0.0125 x 1000 = 19.6g 250 392 = 0.05M = 0.05M 131

Tactics of Chemistry iv. Number of moles of S in 25cm3 If 0.0125 moles are in 250cm3 ? moles are in 25cm3 = 0.0125 x 25 250 = 0.00125 moles v. Determine the mole ratio of R and S Moles of R S 0.00025 0.00125 0.00025 0.00025 1:5 Mole ratio of R : S is 1 : 5 respectively. vi. Write a balanced ionic equation for the reaction between solution R and solution S in the presence of 2M sulphuric (VI) acid. (K=39, Mn = 55, O=16, Fe=56, S=32, N=14, H=1) 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq)+ Mn2+(aq)+ 4H2O(l) Remember. • Working beyond the answer is not allowed. Student may be penalized fully and the working ignored. Working beyond the answer shows that the student is not sure of what is required. • Avoid inconsistency of decimal places like using as 12.0 or 12.00 together. If you chose to use 12.0 use it throughout or if you chose to use 12.00... use it throughout but don`t use values without decimal places. • Rounding off moles is accepted to only 4 significant figures and molarity to 3 significant figures unless it works out exactly to less than the required significant figures. 132

Learning chemistry made easier • Make a tendency of indicating the units and in the correct way. For instance when calculating molarity indicate M after the answer and moles or mol after the answer for moles. Tactical Practice N Question 1 You are provided with:- • Zinc powder, solid S • 0.5M HCl, solution B. • 0.25M NaOH, solution C • Distilled water You are required to determine the : i. Number of moles of hydrochloric acid that remain unreacted. ii. Number of moles of zinc powder that reacted Procedure »» Using a burette, measure 50cm3 of solution B and place it in 100ml beaker. »» Put all of the solid S in the 50cm3 of solution B in the 100ml beaker. Leave the content in the beaker to react for about 5minutes. »» Filter the solution using filter paper and funnel into a 250ml Volumetric flask and top up to the mark with distilled water: Label this solution as solution D. »» Empty the burette and fill it with solution C. Pipette 25cm3 of solution D and place it into an empty 250ml conical flask. Add two drops of methyl orange indicator and titrate solution C against solution D. »» Record the result in the table 1 below. Repeat the titration of solution C against solution D and complete the table 1 below Table 1 Titration I II III (5 marks) Final Burette reading (cm3) Initial Burette reading (cm3) Volume of titre (cm3) a. Calculate the average volume of solution C used. (1mark) b. Calculate the number of moles of: i. Sodium hydroxide used. (1mark) ii. Hydrochloric acid, in 25cm3 of solution D used. (1mark) 133

Tactics of Chemistry iii. Hydrochloric acid in 250cm3 of solution D used. (1mark) iv. Hydrochloric acid in 50cm3 of solution B. (1mark) v. Hydrochloric acid that reacted with Zinc powder. (2marks) vi. Calculate the mass of Zinc that reacted (R.A.M of Zn= 65) (2mark) Confidential In additional to the laboratory apparatus, every student should have • 500ml distilled water • Stopwatch/clock • 250cm3 volumetric flask • Filter paper • 2 Labels • Funnel • 50cm3 burette Reagents • Pipette filler • 2.5g Zinc powder solid S • Pipette • 60cm3 of 0.5M hydrochloric acid, • Two 250cm3 conical flask • White tile Solution B • 100ml beaker • 80cm3 0.25M sodium hydroxide Question 2 You are provided with solution A containing hydrochloric acid. • Solution B containing 5.3 g of anhydrous sodium carbonate Na2CO3 dissolved in 200cm3 distilled water and diluting to form 500cm3 of solution. • Five pieces of metal M ( magnesium ribbon) • You are required to: • Standardize solution A • Determine the rate of reaction between metal M and hydrochloric acid. Procedure 1 Put 10 ml of solution A in a 100ml measuring cylinder and add distilled water to the 100ml mark. Transfer the solution into an empty beaker and label it as solution C. Fill the burette with solution C. Pipette 25cm3 of solution B into a 25ml conical flask. Add 2 drops of methyl orange indicator and titrate with solution C. Record the volume of solution C used in the table. Repeat the procedure twice and com- plete the table below. Titration I II III (4 marks) Final Burette reading (cm3) Initial Burette reading (cm3) Volume of titre (cm3) 134

Learning chemistry made easier a. Determine the average volume of solution C. (1 mark) b. Calculate the number of moles of solution B used. (2 marks) c. Calculate the number of moles of hydrochloric acid in solution C used. (1 mark) d. Calculate the number of moles of hydrochloric acid in 100ml solution C (1 mark) e. Calculate the molar concentration of hydrochloric acid in solution A. (2 marks) Procedure 2. Fill the burette with solution A. Transfer 20cm3 of solution A into an empty 100cm3 beaker. Drop one piece of metal M into the acid in the beaker and start the stop watch immediately. Record the time taken for the reaction to be complete ( i.e no more bubbles being produced ) in the table below. Repeat the procedure using the volume in the table for four more experiments and complete the table. Experiment 1 2 3 45 20 16 12 8 4 Volume of acid (cm3) 0 4 8 12 16 Volume of water (cm3) Concentration of acid (moles/l) Time taken (seconds) a. i. Plot a graph of concentration of acid in moles per litre against time in seconds (3 marks) ii. Explain the shape of your graph.(1 mark) iii. From the graph, determines the rate of reaction at time 10 seconds. (1mark) Confidential In addition to the general apparatus in a chemistry laboratory each student is required to have the following. • 200 ml solution A – 0.2 M HCl • Stopwatch • 100 ml solution B – 0.1 M anhy- • Pipette • Burette drous sodium carbonate. • Complete stand • 5 pieces of Solid M . Magnesium • 2 conical flasks - 250 ml • 100 ml beaker ribbon each 2cm long ( freshly scratched to remove any coating of magnesium oxide). You are provided with: - Question 3 • A d0.i2baMsiscoalcuitdio, Hn2oCf2Oso4d. xiuHm2Ohycdonrotaxiindien(gs1ol2u.6tigoninD1)0. 0cm3 of solution W. • A • You are required to determine the value of X in the formula H2C2O4. x H2O. 135

Tactics of Chemistry Procedure I: Fill the burette with solution W. Using pipette filler and a pipette, transfer 25.0cm3 of solution D in a clean conical flask. Add two drops of phenolphthalein indicator and titrate solution W against solution D. Record your results in table I below. Repeat the titration two more times and complete the table. Retain solution W to be used in procedure II. Table I Titration I II III Final Burette reading (cm3) (4 marks) Initial Burette reading (cm3) Volume of titre (cm3) i. Calculate the average volume of solution W used. (1mark) ii. Calculate the concentration of the acid, solution W in moles per litre. (2marks) iii. Calculate the relative formula mass of the acid. (2marks) iv. Determine the value of X in H2C2O4. x H2O. (2marks) b. You are provided with • Solution U, acidified potassium manganate (VII) solution. • Solution W, oxalic acid solution. You are required to determine how the rate of reaction of acidified potassium manganate (VII) with oxalic acid varies with change in temperature. Procedure II Using a measuring cylinder, place 10.0cm3 portions of solution U into five test tube placed in a test tube rack. Clean the measuring cylinder and use it to place 10.0cm3 of solution W into a boiling tube. Prepare a water bath by placing about 200cm3 of water in a beaker and start heating it. Insert a thermometer in the solution W in the boiling tube in the warm water bath till the solution W attains a temperature of 400C. Remove the boiling tube from the water bath and place it in a test tube rack and add the first portion of solution U and at the same time start a stop watch. Record the time taken for the purple colour of the mixture to decolourise in table II. Repeat the experiment by using 10.0cm3 of solution W at temperatures of 50, 60, 70 and 800C. Record the times taken in seconds in table II below. i. Complete the table by computing 1/t (sec-1) (5 marks) 136

Learning chemistry made easier Table II Temperature of solution W(0C) 40 50 60 70 80 Time for colour to decolourise (seconds) 1/t (sec-1) ii. Plot a graph of 1/t (sec-1) (y axis) against temperature 0C. (3marks) iii. From the graph determine the time taken for decolourization of the mixture, if the temperature of the solution W was 650C. (2marks) iv. How does the rate of reaction of potassium manganate (VII) with oxalic acid vary with temperature? (1 mark) Confidential In addition to the general apparatus in a chemistry laboratory each student is required to have the following. • 200cm3 glass beaker (empty) • About 150cm3 of solution W. • 10cm3 measuring cylinder. • About 100cm3 of 0.2M solution D. • Thermometer (-10 – 1000C) • Burette (0 – 50cm3) • Six clean dry test – tube on a test • Pipette (25cm3) • Pipette filler tube rack. • 3 conical flask • Tripod stand • White tile • Wire gauge • Clamp and stand • Stop watch / clock. • Phenolphthalein indicator with a • Boiling tube -1 dropper • About 60cm3 of solution U. Question 4 You are provided with solutions P and Q. • Solution P is acidified potassium manganate (VII) (KMnO4) containing 3.16g per litre of solution . • Solution Q was prepared by dissolving 4.17g of solid FeSO4.XH2O in distilled water and the solution made to 250cm3 of solution. You are required to determine the value of X in the formula FeSO4.XH2O. Procedure: • Place solution P in a burette. • Pipette 25cm3 of solution Q into a 250cm3 conical flask. • Titrate solution Q with solution P until a permanent pink colour just appears. Record your results in table 1 below. Repeat the above procedure two more times. 137

Tactics of Chemistry Table I I II III Titration Final Burette reading (cm3) Initial Burette reading (cm3) Volume of solution P used (cm3) Calculate; a. the average volume of solution P used (1mark) i. Concentration of potassium manganate (VII) in moles per litre (K = 39, Mn = 55, O=16) (1mark) ii. The number of moles of potassium manganate (VII) used . (1mark) b. Calculate the concentration of solution Q in g/litre. (1mark) c. Given the ionic equation for the above reaction is MnO4- (aq) + 5Fe2+(aq) + 8H+ (aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O (l) Determine the number of moles of FeSO(14.mXHar2Ok)in i. 25cm3 of solution Q ii. 1000cm3 of solution Q (1mark) Determine : formula mass of (1mark) The relative X in the formula FFeeSSOO44..XXHH22OO. (Fe = 56, i. The value of S = 32, O=16, H=1) ii. (1mark) Confidential In addition to the general apparatus in a chemistry laboratory each student is required to have the following. • A bout 75cm3 of solution P • A bout 100cm3 of solution Q • One burette 0 – 50ml • Solution Q. • One pipette 25ml iPnre4p0a0recmd b3 yodf iss2oMlvinsuglp1h6u.7rgicFe(SVOI)4.7aHci2dO, • Two conical flasks 250ml dHi2lSuOti4n(gimtmo eodniaeteliltyreaftoefr weighing) and Notes distilled water. solution using • Solution P This solution is to be prepared in the Prepared by dissolving 3.2g of KMnO4 in morning of the examination & supplied 400cm3 of 2M sulphuric (VI) acid, H2SO4 to candidates in containers sealed with acid and diluting to one litre solution aluminium foil. using distilled water. 138