C16 – Solutions 19 Fifth Edition C16. Solutions TABLE OF CONTENTS C16. Solutions 19 C16.1 Vapour Pressure and Raoult’s Law ………………………………………………………………………………………..……20 C16.2 Colligative Properties …………………………………………………………………………………………………………...……26 C16.3 Abnormal Molar Mass and van ’t Hoff Factor ………………………………………………………………………………29 Test Practice Problems ……………………………………………………………………………………………………………...…………32 Answers Key …………………………………………………………………………………………………………………………….………….37
C16 – Solutions 20 C16.1 Vapour Pressure and Raoult’s Law CONCEPTS 1. Solution, solvent and solute 2. Solubility and the effect of temperature and pressure on it 3. Vapour pressure 4. Raoult’s Law for solutions with volatile and non-volatile components 5. Ideal and Non-ideal solutions 6. Positive and negative deviations from Raoult’s Law in non-ideal solutions PRE-READING You may refer to the following source: Category Book Name (Edition) Chapter Section REQUIRED NCERT XII (Part I) 2 2.1, 2.2, 2.3, 2.4, 2.5 SYNOPSIS 1. Column II Column I a) Mass of solute component in solution in kilograms × 100 I. Mass percentage II. Volume percentage Total mass of the solution in kilograms III. Mole fraction IV. Molality b) Volume of the solute component in solution in litres × 100 V. Molarity Total volume of solution in litres VI. Parts per million VII. Mass by volume percentage c) Number of moles of a component Total number of moles of all the components d) Number of moles of solute component Mass of the solvent in kilograms e) Number of moles of solute component. Volume of solution in litres f) Number of parts of the component × 106 Total number of parts of all components g) Mass of the solute in grams in solution × 100 Total volume of solution in mililitres 2. ������������×������������ = ������������×������������ Molarity equation ������������ ������������ Normality equation ������������ × ������������ = ������������ × ������������ C16.1
C16 – Solutions 21 3. Factors affecting the solubility of a solid in liquid. I. ������������������������������������������������������������������↗↘foforreenxdootthheerrmmicicddisisssooluluttioionn→→tteemmppeerraattuurree↓↑ssoolulubbiliiltityy↓↑ II. Nature of solute & solvent → like dissolves like. 4. Factors affecting the solubility of a gas in liquid. I. Temperature → solubility decreases with rise in temperature. II. Nature of gas & solvent III. Pressure – with increase in pressure solubility increases 5. Henry’s law ������������ = ������������������������ or ������������ = ������������ × ������������. 6. a) The total vapour pressure of a binary solution containing two volatile liquids ������ and ������ is given by ������������������������������������ = ������������° + (������������° − ������������°)������������. ������������ ������������+������������ b) Now if ������������ and ������������ represent the mole fraction of the components in vapour phase then ������������ = and ������������ = ������������ . ������������+������������ 7. Diagrams a) Ideal Solution b) Positive deviation c) Negative deviation
C16 – Solutions 22 8. PRE-READING EXERCISE Q1. Match the column A) Mole fraction of a P) Total Number of parts of the component solution × 106 component number of parts of all components of the B) Mass percentage of a Q) Mass of the component in the solution × 100 component (������/������) Total mass of the solution C) Volume percentage of a Number of moles of the component component (������/������) R) Total number of moles of all the components D) Parts per million of a S) Volume of the component component Total volume of solution × 100 E) Molality (������) Moles of solute T) Mass of solvent in ������������ F) Molarity U) Moles of solute Volume of solution in ������������������������������������ C16.1
C16 – Solutions 23 Q2. For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its ___________ present in solution. Q3. The solutions which obey Raoult’s law over the entire range of concentration are known as ___________. Q4. The solubility of a gas in a liquid ___________ (increases/decreases) with increase in temperature and ___________ (increases/decreases) with increase in pressure. Q5. For a non-ideal solution, the vapour pressure of such a solution is either higher or lower than that predicted by ___________. Q6. For an ideal solution, the change in enthalpy of mixing of the pure components to form the solution is ___________ and the change in volume of mixing is ___________. IN-CLASS EXERCISE LEVEL 1 Q1. Consider the following graph plotted between the vapour pressure of two volatile liquids against their respective mole fraction and select the correct option. ������������ = 1 ������������ = 0 ������������ = 0 ������������ = 1 Mole fraction A) When ������������ = 1 and ������������ = 0, then ������ = ���������0��� B) When ������������ = 1 and ������������ = 0, then ������ = ���������0��� C) When ������������ = 0 and ������������ = 1 then ������ < ���������0��� D) When ������������ = 0 and ������������ = 1 then ������ = ���������0��� Q2. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 ������������ ������������ at 300 ������. The vapour pressure of pure propyl alcohol is 200 ������������ ������������. If the mole fraction of ethyl alcohol is 0.6, what will be the value of its vapour pressure (in ������������ ������������) at the same temperature? Q3. The azeotropic mixture of water and ������2������5������������ boils at 78.15°������. When this mixture is distilled, it is possible to obtain: A) Pure ������2������ B) Pure ������5������5������������ C) Pure ������2������ as well as ������5������5������������ D) neither ������2������ nor ������5������5������������ in their pure state Q4. Equal masses of methane and oxygen are mixed in an empty container at 25°������. What is the fraction of the total pressure exerted by oxygen?
C16 – Solutions 24 LEVEL 2 Q5. The process of two components forming a solution can be represented as three step-wise reactions where reactants and products are: Reaction No. Reactant Product Reaction Enthalpy I. Pure solvent Separated solvent molecules ∆������1 II. Pure solute Separated solute molecules ∆������2 III. Separated solvent Solute molecules ∆������3 Solution so formed will be ideal if A) ∆������������������������������ = ∆������1 − ∆������2 − ∆������3 B) ∆������������������������������ = ∆������3 − ∆������1 − ∆������2 C) ∆������������������������������ = ∆������1 + ∆������2 + ∆������3 D) ∆������������������������������ = ∆������1 + ∆������2 − ∆������3 Q6. Two liquids ������ and ������ form an ideal solution. The mixture has a vapour pressure of 400 ������������ ������������ at 300 ������ when mixed in the molar ratio of 1: 1 and a vapour pressure of 350 ������������ ������������ when mixed in the molar ratio of 1: 2 at the same temperature. What will be the values of the vapour pressures of the two pure liquids ������ and ������? Q7. Two liquids ������ and ������ form an ideal solution at 300 ������, vapour pressure of the solution containing 1 ������������������ of ������ and 3 ������������������ of ������ is 550 ������������ ������������. At the same temperature, if 1 ������������������ of ������ is further added to this solution, vapour pressure of the solution increases by 10 ������������ ������������. What will be the vapour pressure (in ������������ ������������) of ������ and ������ in their pure states? Q8. A non-ideal solution was prepared by mixing 30 ������������ chloroform and 50 ������������ acetone. The volume of mixture will be A) > 80 ������������ B) < 80 ������������ C) = 80 ������������ D) ≥ 80 ������������ Q9. A binary liquid solution is prepared by mixing ������ −heptane and ethanol. Which one of the following statements is correct regarding the behavior of the solution? A) The solution formed is an ideal solution B) The solution is non-ideal, showing positive deviation from Raoult’s law C) The solution is non-ideal, showing negative deviation from Raoult’s law D) ������ −heptane shows positive deviation while ethanol show negative deviation from Raoult’s law LEVEL 3 Q10. The vapour pressure of pure liquids ������ and ������ are 400 and 800 ������������ ������������ respectively, at 350������ . Find out the composition of the liquid mixture if total vapour pressure is 500 ������������ ������������. Also find the composition of the vapour phase. HOMEWORK LEVEL 1 Q1. Vapour pressure of ′������′ is 70 ������������ of ������������ at 25° ������. It forms an ideal solution with ′������′ in which mole fraction of ������ is 0.8. If the vapour pressure of the solution is 84 ������������ of ������������ at 25° ������, what is the vapour pressure of pure ′������′ at 25°������? Q2. Which one of the following is not correct for an ideal solution? A) ∆������ = 0 B) ∆������ = 0 C) Both ������ and ������ D) ∆������ = ∆������ ≠ 0 C16.1
C16 – Solutions 25 LEVEL 2 Q3. The vapour pressure of two pure liquids ������ and ������ are 80 and 60 ������������������������ respectively. What will be the total vapour pressure of the ideal solution obtained by mixing 3 moles of ������ and 2 moles of ������?’ Q4. At 25° ������, the total pressure of an ideal solution obtained by mixing 3 ������������������������������ of ′������′ and 2 ������������������������������ of ′������′ is 184 ������������������������. What is the vapour pressure (in ������������������������) of pure ′������′ at the same temperature? (vapour pressure of pure ′������′ at 25° ������ is 200 ������������������������) Q5. Which of these will form a maximum boiling azeotrope? A) ������6������6 + ������6������5������������3 solution B) ������������������3 + ������2������ solution C) ������2������5������������ + ������2������ solution D) None of these Q6. The LPG cylinders used for cooking in Indian households generally have two liquefied gases, propane and butane as its components forming an ideal solution together. Given that the total vapour pressure of a binary solution is 500 ������������ ������������, partial vapour pressure of pure liquefied propane is 600 ������������ ������������, and the molar ratio of propane to butane is 2: 1, find the composition of the vapour phase in the cylinder. Q7. Benzene and toluene form nearly ideal solutions. At 20° ������, the vapour pressure of pure benzene is 75 ������������������������ and that of pure toluene is 21 ������������������������. What is the partial vapour pressure of 78 ������ of benzene and 46 ������ of toluene? Q8. 1 ������������������������ of liquid ������ and 2 moles of liquid ������ make a solution having a total vapour pressure of 38 ������������������������. The vapour pressures of pure ������ and pure ������ are 45 ������������������������ and 36 ������������������������ respectively. Is the described solution ideal in nature? If not, what is the kind of deviation that it shows? LEVEL 3 Q9. A binary solution of ������ and ������ is formed such that the total vapour pressure of the solution is equal to 800 ������������ ������������. Also, it is given that the vapour pressure of pure component ������ is thrice that of the vapour pressure of pure ������. Find out the mole fraction of ������ (������������) in the initial solution given that the total vapour pressure becomes 600 ������������ ������������ if the composition is changed such that the new mole fraction of ������ = 2 × ������������.
C16 – Solutions 26 C16.2 Colligative Properties CONCEPTS 1. Colligative properties of solutions 2. Relative lowering of vapour pressure for solutions 3. Elevation in boiling point for dilute solutions 4. Depression in freezing point for dilute solutions 5. Osmotic pressure as a colligative property PRE-READING You may refer to the following source: Category Book Name (Edition) Chapter Section 2 2.6 REQUIRED NCERT XII (Part I) SYNOPSIS 1. Colligative Properties Formula Formula for molar mass of solute (������������) Relative lowering of vapour pressure ������0 − ������������ = ������2 ������2 = ������2 × ������1 × ������0 ������0 ������1 − ������0 ������������ Osmotic pressure ������ = ������������������ ������2 = ������2������������ ������������ Elevation in boiling point ∆������������ = ������������ × ������ ������2 = 100 × ������������ × ������2 ∆������������ × ������1 Depression in freezing point ∆������������ = ������������ × ������ ������2 = 1000 × ������������ × ������2 ∆������������ × ������1 Here ������0 = Vapour pressure of pure component ������������ = Vapour pressure of the solution ������2= Mole fraction of solute ������ = Osmotic pressure ������ = Concentration of solution ������= Universal Gas Constant ������ = Absolute temperature ∆������������ = Elevation in boiling point ������������ = Ebulloscopic constant ������= Molality ∆������������= Depression in Freezing point ������������ = Cryoscopic constant ������1 = Molar mass of solvent ������2 = Molar mass of solute ������1 = Given weight of solvent ������2 = Given weight of solute C16.2
C16 – Solutions 27 2. a) A solution having same osmotic pressure at a given temperature as that of a given solution: ISOTONIC. b) A solution whose osmotic pressure is less than that of another: HYPOTONIC. c) Solution with two components: BINARY. d) A solution which contains maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature: SATURATED. e) A solution whose osmotic pressure is more than that of another: HYPERTONIC. f) A solution where solid and solvent both are in solid phase: ALLOY PRE-READING EXERCISE Q1. ___________ is the spontaneous net movement of solvent molecules through a semi-permeable membrane into a region of higher solute concentration. Q2. ___________ is the minimum pressure which needs to be applied on a solution to prevent the inward flow of water across a semipermeable membrane. Q3. Two solutions that have the same osmotic pressure at a given temperature are called ___________. Q4. The lowering of ___________ pressure of a solution on addition of a non-volatile solute causes a lowering of the freezing point compared to that of the pure solvent. Q5. Any liquid boils at the temperature at which its vapour pressure is equal to the ___________. IN-CLASS EXERCISE LEVEL 1 Q1. The molal freezing point constant for water is 1.86 ������ ������������ ������������������−1. If 342 ������ cane sugar (������12������22������11) are dissolved in 1000 ������ of water, at what temperature the solution will freeze? Q2. When 10 ������ of a non-volatile solute is dissolved in 100 ������ of benzene, it raises boiling point by 1°������, then molecular mass of the solute is (������������ for ������6������6 = 2.53 ������������������������������������−1 ): Q3. Osmotic pressure is 0.0821 ������������������ at temperature of 300 ������. Find the concentration in moles per litre. Q4. The average osmotic pressure of human blood is 7.8 ������������������ at 37°������. What is the concentration of an aqueous sugar solution that could be used in the blood stream? LEVEL 2 Q5. The osmotic pressure of 0.4% urea solution is 1.66 ������������������ and that of a solution of sugar of 3.42% is 2.46 ������������������. What will be the osmotic pressure of the resultant solution when both the solutions are equally mixed? Q6. A substance has a boiling point of 117°������ and freezing point of −13°������; ������������ and ������������ are 3.6 and 3.0 ������/������������������������������. What is the ratio of its enthalpy of vaporization to that of its fusion? Q7. A solution containing 1.8������ of a compound (empirical formula ������������2������) in 40 ������ of water is observed to freeze at −0.465°������. What will be the molecular formula of the compound? (������������ of water = 1.86 ������������������������������������−1) Q8. At 25℃ a 5% weight by volume aqueous solution of glucose (molecular weight = 180 ������ ������������������−1) is isotonic with 2% weight by volume aqueous solution containing an unknown solute. What is the molar mass of the unknown solute?
C16 – Solutions 28 LEVEL 3 Q9. Dry air is passed through a solution containing 10 ������ of the solute in 90 ������ of water and then through pure water. The loss in weight of solution is 2.5 ������ and that of pure solvent is 0.05 ������. Calculate the molecular weight of the solute. Q10. A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 65 ������������3 of benzene (density= 0.8 ������������������−3). At room temperature vapour pressure of this solution is 98 ������������ ������������ while that of pure benzene is 100 ������������ ������������. Calculate the molality of this solution. If the freezing temperature of this solution is 0.72 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene? HOMEWORK LEVEL 1 Q1. In countries close to the polar region, the roads are sprinkled with ������������������������2. This is A) To remove the ice from roads B) To reduce the snow fall C) To minimize pollution D) To minimize the accumulation of dust on the road. Q2. If 20 ������ of a solute was dissolved in 500 ������������ ������f water and osmotic pressure of the solution was found to be 600 ������������ of ������������ at 15°������, then what is the molecular weight of the solute? Q3. If 0.15 ������ of a solute dissolved in 15 ������ of solvent boils at a temperature higher by 0.216°������ than that of the pure solvent, what is the molecular weight of the substance (molal elevation constant for the solvent is 2.16 ������������������������������������−1) ? LEVEL 2 Q4. What is the osmotic pressure of a 5% (������/������) solution of cane sugar (������12������22������11) at 69°������? Q5. What should be the amount of solute (molar mass 60 ������������������������−1) that must be added to 180 ������ of water so that the vapour pressure of water is lowered by 10% ? Q6. In 100 ������ of naphthalene, 2.16 ������ of sulphur was dissolved. Melting point of naphthalene = 80°������ . ∆������������ = 0.6°������, ������������ = 35 ������������������/������ of naphthalene. What is the molecular formula of sulphur added? A) ������2 B) ������8 C) ������6 D) ������4 Q7. A 5% solution of cane sugar (molar mass 342 ������/������������������) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in ������/������������������ is A) 136.2 B) 171.2 C) 68.4 D) 34.2 LEVEL 3 Q8. A solution of a non- volatile solute in water freezes at − 30°������. The vapour pressure of pure water at 298 ������ is 23.51 ������������ ������������ and ������������ for water is 1.86 ������/������������������������������. Calculate the vapor pressure of this solution at 298 ������. Q9. A depression in freezing point of 5.58℃ is observed for an aqueous solution of ������. Given that the molal freezing constant of water is 1.86 ������������������/������������������, find if the given solution is isotonic/hypertonic/hypotonic with respect to a solution of 3������ glucose. (Given that the density of 3������ glucose solution is 1.2 ������/������������) C16.2
C16 – Solutions 29 C16.3 Abnormal Molar Mass and van ’t Hoff Factor CONCEPTS 1. Abnormal molar mass and van ’t Hoff factor 2. Relation of degree of dissociation and degree of association of solute to the van ’t Hoff factor 3. Modified equations including van ’t Hoff factor for colligative properties PRE-READING You may refer to one of the following sources: Category Book Name (Edition) Chapter Section REQUIRED NCERT XII (Part I) 2 2.7 1. Van’t Hoff factor SYNOPSIS a) ������ = 0 Behaviour b) ������ = 1 1. Impossible c) ������ < 1 2. Neither association nor dissociation d) ������ > 1 3. Association 4. Dissociation 2. Van’t Hoff factor observed colligative property ������ = calculated colligative property Normal molar mass ������ = Observed or abnormal molar mass Total number of moles of particles after association/dissociation ������ = Number of moles of particles before association/dissociation 3. Modified expressions for substances undergoing Colligative property association or dissociation. Relative lowering of V.P. ������������ − ������������ = ������������������ Osmotic pressure ������������ Elevation in boiling point Depression in freezing point ������ = ������������������������ ∆������������ = ������������������������ ∆������������ = ������������������������
C16 – Solutions 30 4. In terms of molar masses ������������ In terms of ������ (Van’t Hoff factor) and Parameter (calculated molar mass) and ������������ ������ (number of particles) Degree of Dissociation (observed molar mass) ������ − 1 Degree of association ������ = ������ − 1 ������ = ������������ − ������0 ������0(������ − 1) ������ ������ = 1 − ������ × ������ − 1 ������ = ������0 − ������������ . ������ ������ 1 ������0 − PRE-READING EXERCISE Q1. When there is dissociation of solute into ions, the experimentally determined molar mass is always ___________ than the true value. (higher/lower) Q2. Molecules of ethanoic acid ___________ in benzene. Q3. In case of association, value of van ’t Hoff factor is ___________ than unity. (less/greater) Q4. The value of van ’t Hoff factor for ������������������ (assuming complete dissociation) will be ___________. IN-CLASS EXERCISE LEVEL 1 Q1. If sodium sulphate is considered to be completely dissociated into cations and anions in an aqueous solution, what will be the change in freezing point of water (∆������������), when 0.01 ������������������������ of sodium sulphate is dissolved in 1 ������������ of water? (������������ = 1.86 ������������������������������������−1) Q2. When 20 ������ of naphthoic acid (������11������8������2) is dissolved in 50 ������ of benzene (������������ = 1.72 ������������������������������������−1), a freezing point depression of 2 ������ is observed. What will be the van ’t Hoff factor (������) ? Q3. If ������ is the degree of dissociation of ������������2������������4 and ������ is the van ’t Hoff factor, find the relation between ������ and ������. LEVEL 2 Q4. 0.004 ������ ������������2������������4 is isotonic with 0.01 ������ glucose. Find the degree of dissociation of ������������2������������4? Q5. A 0.025 ������ solution of a monobasic acid has a freezing point of −0.06℃, find the value of ������������ of this acid. ������������ of water = 1.86 ������/������������������������������. Assume that molarity is equal to molality. Q6. Vapour pressure of benzene at 30℃ is 160 ������������ ������������ ������������. When in 2.4 ������������������ of benzene, 6 ������ of acetic acid was dissolved, the solution has a vapour pressure of 158 ������������ ������������ ������������. Given that molar mass of acetic acid is 60 ������/������������������. Calculate: I. van ’t Hoff factor II. The degree of association of acetic acid in benzene at 30℃. Q7. The ������������ of 1 ������ solution of a weak monobasic acid (������������) is 2. What is the van’t Hoff factor for this solution? Q8. Following are equimolal aqueous solutions (Assume Molarity = Molality): I. 1 ������ urea II. 1 ������ ������������������ III. 1 ������ ������������������������2 IV. 1 ������ ������������3������������4 Arrange them in the order of increasing ������) Boiling point ������������) Freezing point ������������������) Osmotic pressure and ������������) Vapour pressure assuming complete ionization for the given ionic salts. C16.3
C16 – Solutions 31 LEVEL 3 Q9. ������ ������ of non-electrolyte compound (molar mass = 200 ������/������������������) is dissolved in 1.0 ������������������������������ of 0.05 ������ ������������������������ aqueous solution. The osmotic pressure of this solution is found to be 4.1 ������������������ at 27° ������. Calculate the value of ������. Assume complete dissociation of ������������������������ and ideal behavior of this solution. (������ = 0.082 ������������������������������ ������������������ ������������������−1 ������−1). HOMEWORK LEVEL 1 Q1. Observe the following abbreviations ������������������������ = Observed colligative property ������������������������ = Theoretical colligative property assuming normal behavior of solute. Now, van ’t Hoff factor (������) is given by A) ������ = ������������������������ × ������������������������ B) ������ = ������������������������ + ������������������������ C) ������ = ������������������������ − ������������������������ D) ������ = ������������������������ ������������������������ Q2. An aqueous solution containing an ionic salt having molality equal to 0.2 freezes at −0.704℃ . The van ’t Hoff factor of the ionic salt is (Given ������������ for water = 0.86 ������/������) A) 3 B) 2 C) 4 D) 5 LEVEL 2 Q3. Acetic acid associates in benzene to form a dimer. 1.65 ������ of acetic acid when dissolved in 100 ������ of benzene raised the ������. ������. by 0.36° ������. Calculate the van ’t Hoff factor and the degree of association of acetic acid. (������������ for benzene = 2.57℃/������) Q4. The degree of dissociation (������) of a weak electrolyte ������������������������ is related to van ’t Hoff factor (������) by the expression A) ������ = ������−1 B) ������ = ������−1 C) ������ = ������+������−1 D) ������ = ������+������+1 (������+������−1) ������+������+1 ������−1 ������−1 Q5. A 0.1 ������ solution of potassium sulphate, ������2������������4 is ionized to the extent of 90%. What would be its osmotic pressure at 27℃? Q6. When dissolved in benzene, a compound ������38������30 partially dissociates by the following equation: ������38������30 ⇌ 2������19������15 25.6 ������ of ������38������30 is dissolved in 400 ������ of benzene, the freezing point is lowered by 0.680°������. What percentage of ������38������30 molecules have dissociated? (������������ = 4.9 ������������ ������ ������������������−1) Q7. What is the degree of dissociation ′������′, of a weak electrolyte (which dissociates into ������ particles)in terms of ������ and ������? Q8. The van ’t Hoff factor for a 0.1 ������ ������������2(������������4)3 solution is 4.20. What is the degree of dissociation of ������������2(������������4)3? LEVEL 3 Q9. The degree of dissociation of ������������(������������3)2 in a dilute aqueous solution, containing 7.0 ������ of the salt per 100 ������ of solution at 100℃ is 70 percent. If the vapour pressure of water at 100℃ is 760 ������������ ������f ������������, calculate the vapour pressure of the solution. (Molar mass of ������������(������������3)2 = 164 ������/������������������)
C16 – Solutions 32 Test Practice Problems Purpose: To practice a mixed bag of questions in a speed based format similar to what you will face in entrance examinations. In most entrance examinations, you will get not more than 3 minutes to attempt a question. Hence, you need to be able to attempt a question in less than 3 minutes, and at the end of 3 minutes skip the question and move to the next one. Approach: Attempt the Test Practice Problems only when you have the stipulated time available at a stretch. Start a timer and attempt the section as a test. DO NOT look at the answer key / solutions after each question. DO NOT guess a question if you do not know it. Competitive examinations have negative marking. Solve as much as possible within the stipulated time, and then fill the OMR provided at the end of the TPP. Fill the table at the end of the TPP and evaluate the number of attempts, and accuracy of attempts, which will help you evaluate your preparedness level for the chapter. Q1. The vapour pressure of a pure liquid ������ is 40 ������������ ������������ at 310 ������. The vapour pressure of this liquid in a solution with liquid ������ is 32 ������������ ������������. What is the mole fraction of ������ in the solution if it obeys Raoult’s law? A) 0.2 B) 0.75 C) 0.25 D) 0.8 Q2. Which of the following liquid pairs shows a positive deviation from Raoult’s law? A) Water-hydrochloric acid B) Benzene-methanol C) Water-nitric acid D) Acetone-chloroform Q3. Assertion (A): An Azeotropic liquid mixture is one that boils with unchanged composition in solution and in vapour state. Reason (R): A maximum boiling azeotrope is formed due to large positive deviation from Raoult’s Law. A) Both A and R are correct and R is the correct explanation for A B) Both A and R are correct but R is not the correct explanation of A C) A is correct but R is incorrect D) A is incorrect but R is correct Q4. In a mixture of ������ and ������. Components show negative deviation when A) ������ − ������ interaction is stronger than ������ − ������ and ������ − ������ interaction B) ������ − ������ interaction is weaker than ������ − ������ and ������ − ������ interaction C) ∆������������������������ > 0 D) ∆������������������������ = 0 Q5. An ideal solution is formed on mixing heptane and octane. At 373 ������, the vapour pressures of the two liquid components (heptane and octane) in their pure form are 105 ������������������ and 45 ������������������ respectively. What will be the vapour pressure of the solution obtained by mixing 25 ������ of heptane and 35 ������ of octane? (molar mass of heptane = 100 ������������������������−1 and of octane = 114 ������������������������−1). A) 82.5 ������������������ B) 75 ������������������ C) 70 ������������������ D) 72 ������������������ Q6. Ethylene glycol is used as an antifreeze in cold climate. What will be the minimum mass of ethylene glycol which should be added to 4 ������������ of water to prevent it from freezing at −6°������ ?(������������ for water = 1.86 ������������������ ������������������−1, molar mass of ethylene glycol = 62������������������������−1) A) 500 ������ B) 400 ������ C) 800 ������ D) 1000 ������ T.P.P.
C16 – Solutions 33 Q7. The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. What is the molality of the solution? A) 0.70 ������ B) 0.45 ������ C) 1.43 ������ D) 0.90 ������ Q8. 200 ������������ of an aqueous solution of a protein contains its 1.26 ������. The osmotic pressure of this solution at 300 ������ is found to be 2.57 × 10−3 ������������������. The molar mass of protein will be (������ = 0.083 ������ ������������������ ������������������−1������−1): A) 51022 ������������������������−1 B) 122044 ������������������������−1 C) 31011 ������������������������−1 D) 61038 ������������������������−1 Q9. Two aqueous solutions, ������ and ������, are separated by a semi-permeable membrane. The osmotic pressure of solution ������ immediately begins to decrease. Which of the following statement is true? A) The solvent molecules are moving from the solution of higher osmotic pressure to that of lower osmotic pressure B) The real osmotic pressure of solution ������ is greater than that of solution ������ C) Solvent molecules are moving from solution ������ into solution ������ D) Both ������ and ������ are true statements Q10. A 6% (������/������) solution of urea is isotonic with B) 0.05 ������ solution of glucose D) 25% (������/������) solution of glucose A) 1 ������ solution of glucose C) 6% (������/������) solution of glucose Q11. For a weak monobasic acid, if ������������������ = 5, then at a concentration of 0.1������ of the acid solution, the van ’t Hoff factor is A) 1.01 B) 1.02 C) 1.10 D) 1.20 Q12. The freezing point of water is depressed by 0.37℃ in a 0.01 ������������������������������ ������������������������ solution. The freezing point of 0.02 ������������������������������ solution of urea is depressed by A) 0.37°������ B) 0.74°������ C) 0.185°������ D) 0°������ Q13. The order of osmotic pressure of isomolar solution of ������������������������2, ������������������������ and sucrose is A) ������������������������2 > ������������������������ >sucrose B) ������������������������ > ������������������������2 > sucrose C) sucrose > ������������������������ > ������������������������2 D) ������������������������2 > sucrose > ������������������������ Q14. Which one of the following aqueous solutions will exhibit highest boiling point assuming complete ionization of ionic salts? A) 0.01 ������ ������������2������������4 B) 0.01 ������ ������������������3 C) 0.04 ������ urea D) 0.02 ������ glucose Q15. The freezing point of one molal ������������������������ solution assuming ������������������������ to be 100 % dissociated in water is (molal depression constant = 1.86 ������������������������������������−1): A) −2.72°������ B) −3.72°������ C) 2.72°������ D) 3.72°������ Q16. Solution ������ contains 7������/������ ������������������������2 and solution ������ contains 7 ������/������ of ������������������������ . Given that molar mass of ������������������������2 = 95 ������/������������������ and ������������������������ = 58.5 ������/������������������ , at room temperature, the osmotic pressure of A) Solution ������ is greater than that of ������ B) Both have same osmotic pressure C) Solution ������ is greater than that of ������ D) Can’t determine Q17. An aqueous solution of urea is found to boil at 100.52℃. Given ������������ for water is 0.52 ������ ������������ ������������������−1, the mole fraction of urea in the solution is A) 1 B) 0.5 C) 0.018 D) 0.25
C16 – Solutions 34 Q18. A solution of urea (mol. mass 56 ������ ������������������−1) boils at 100.18° ������ at the atmospheric pressure. If ������������ and ������������ for water are 1.86 and 0.512 ������������������������������������−1 respectively. The above solution will freeze at A) −6.54℃ B) −0.654℃ C) 6.54℃ D) 0.654℃ Q19. ������������ for water is 1.86 ������ ������������ ������������������−1. If your automobile radiator holds 1.0 ������������ of water, how many grams of ethylene glycol (������2������6������2) must you add so that the water does not freeze even at −2.79℃? A) 27 ������ B) 72 ������ C) 93 ������ D) 39 ������ Q20. The van ’t Hoff factor for ������������������������2 at 0.01 ������ concentration is 1.98. The percentage dissociation of ������������������������2 at this concentration is A) 49 B) 69 C) 89 D) 98 Q21. The freezing point depression constant for water is −1.86°������ ������−1. If 5.00 ������ ������������2������������4 is dissolved in 45.0 ������ ������2������, the freezing point is changed by −3.72℃. calculate the van ’t Hoff factor for ������������2������������4 A) 0.381 B) 2.05 C) 2.56 D) 3.11 Q22. A 0.1 molal aqueous solution of a weak monobasic acid is 30% ionized. If ������������ for water is 1.86℃/������, the freezing point of the solution will be A) −0.18℃ B) −0.54℃ C) −0.36℃ D) −0.24℃ Q23. At a certain Hill station, water boils at 96℃. The amount of ������������������������ that should be added to one litre of water so that it boils at 100℃ will be (������������for ������2������ = 0.52 ������/������) A) 450 ������ B) 225 ������ C) 125 ������ D) 250 ������ For questions 24 and 25 Each question contains STATEMENT-1 (assertion) and STATEMENT -2 (Reason). Each question has 4 choices (A), (B), (C), and (D) out of which ONLY ONE is correct. Choose the correct option as under: A) Statement-1 true, statement-2 is true; statement-2 is a correct explanation of statement-1. B) Statement-1 true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1. C) Statement-1 true, statement-2 false. D) Statement-1 is false, statement-2 is true. Q24. Statement-1: If on mixing the two liquids, the solution becomes hot, it implies that it shows negative deviation from Raoult’s law. Statement -2: Solutions which show negative deviation are accompanied by decrease in volume. Q25. Statement-1: Vapour pressure of water is less than 1.013 bar at 373 ������. Statement-2; Water boils at 373 ������ as the vapour pressure at this temperature becomes equal to atmosphere pressure. For questions 26-28 In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as A) If both assertion and reason are true, and reason is the true explanation of the assertion. B) If both assertion and reason are true, but reason is not the true explanation of the assertion. C) If assertion is true, but reason is false. D) If both assertion and reason are false. Q26. Assertion: Cooking time is reduced in pressure cooker. Reason: Boiling point inside the pressure cooker is raised. T.P.P.
C16 – Solutions 35 Q27. Assertion: Van ’t Hoff factor for benzoic acid in benzene is less than one. Reason: Benzoic acid behaves as a weak electrolyte in benzene. Q28. Assertion: 0.1 ������ solution of glucose has higher increment in the freezing point than 0.1 ������ solution of urea. Reason: Both solutions have the same ������������ value. Q29. An azeotropic solution of two liquids has boiling point lower than either of them when it A) Shows 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 Q30. A solution of two liquids boils at a temperature more than the boiling point of either of them. Hence, the binary solution shows A) Negative deviation from Raoult’s law B) Positive deviation from Raoult’s law C) No deviation from Raoult’s law D) Positive or negative deviation from Raoult’s law depending upon the composition Q31. Which one of the following aqueous 1 ������������������������������ solutions has the highest freezing point? A) Urea B) Barium chloride C) Potassium bromide D) Aluminium sulphate Q32. If for a sucrose solution elevation in boiling point is 0.1℃ then what will be the elevation in boiling point of ������������������������ solution for the same molal concentration? A) 0.1° ������ B) 0.2° ������ C) 0.05° ������ D) 0.15° ������ Q33. The empirical formula of a non-electrolyte is ������������2������. A solution containing 3 ������ of the compound in 1������ of water exerts the same osmotic pressure as that of 0.05 ������ glucose solution. What is the molecular formula of the compound? A) ������������2������2 B) ������2������4������2 C) ������4������8������2 D) ������3������6������3 Q34. At 80° ������, the vapour pressure of pure liquid ′������′ is 520 ������������ ������������ and that of pure liquid ′������′ is 1000 ������������ ������������. If a mixture solution of ′������′ and ′������′ boils at 80℃ and 1 ������������������ pressure, the amount of ′������′ in the mixture is (1 ������������������ = 760 ������������ ������������) A) 52 mole percent B) 34 mole percent C) 48 mole percent D) 50 mole percent
C16 – Solutions 36 DATA ANALYSIS Guide A # of questions Total problems in TPP B # Attempts Total attempts in OMR C # Correct Total questions correct D # Incorrect Out of the ones marked in OMR E # Unattempted ������ − ������ F Percentage attempts G Percentage Accuracy ������ ������ × 100 ������ ������ × 100 Question type # Correct (C) # Incorrect (I) # Unattempted (U) Easy Medium Hard Tip: To begin with, your accuracy must be high, typically > 60%. Percentage attempts should be > 50% As time progresses, your percentage attempts should increase without a reduction in accuracy. Additionally, you should be able to get > 80% Easy questions correct, as they involve basic recall of the concepts and formulae of the chapter. T.P.P.
C16 – Solutions 37 Answers Key C16.1 VAPOUR PRESSURE AND RAOULT’S LAW PRE-READING EXERCISE Q9. B Q1. A-R, B-Q, C-S, D-P, E- T, F-U LEVEL 3 Q2. Mole fraction Q10. ������: ������ = 3: 1(in solution), ������: ������ = 3: 2 (in vapour phase) Q3. Ideal solution Q4. Decreases, increases HOMEWORK Q5. Raoult’s law LEVEL 1 Q6. Zero, zero Q1. 140 ������������ IN CLASS EXERCISE Q2. D LEVEL 1 LEVEL 2 Q1. B Q3. 72 ������������������������ Q2. 350 ������������ ������������ Q4. 160 ������������������������ Q3. D Q5. B Q4. 1/3 Q6. Propane : butane = 4 ∶ 1 Q7. 50 ������������������������, 7 ������������������������ LEVEL 2 Q8. Non ideal, negative deviation Q5. C LEVEL 3 Q6. 550 ������������ ������������, 250 ������������ ������������ Q7. 400, 600 Q9. 0.3 Q8. B C16.2 COLLIGATIVE PROPERTIES PRE-READING EXERCISE LEVEL 3 Q1. Osmosis Q9. 100������/������������������ Q2. Osmotic pressure Q10. 0.205 ������, 3.512 ������������������������������������−1 Q3. Isotonic solutions Q4. Vapour HOMEWORK Q5. Atmospheric Pressure LEVEL 1 IN CLASS EXERCISE Q1. A Q2. 1200 ������/������������������ LEVEL 1 Q3. 100 ������/������������������ Q1. −1.86°������ LEVEL 2 Q2. 253 ������ ������������������−1 Q3. 3.33 × 10−3 Q4. 4.1 ������������������ Q4. 0.31 ������������������/������ Q5. 60 ������ Q6. B LEVEL 2 Q7. C Q5. 2.06 ������������������ LEVEL 3 Q6. 2.7 Q7. ������6������12������6 Q8. 23.44 ������������ ������������ Q8. 72 Q9. Hypotonic
C16 – Solutions 38 C16.3 ABNORMAL MOLAR MASS AND VAN ’T HOFF FACTOR Ans. PRE-READING EXERCISE iii) I < II < III < IV iv) IV < III < II < I Q1. Lower Q2. Dimerize LEVEL 3 Q3. Less Q9. 13.33������ Q4. Two HOMEWORK IN CLASS EXERCISE LEVEL 1 Q1. D LEVEL 1 Q2. C Q1. 0.0558 ������ LEVEL 2 Q2. 0.5 Q3. 0.512, 0.97 Q3. ������ = (1 + 2������) Q4. A Q5. 6.9 ������������������ LEVEL 2 Q6. 5.38% Q7. ������ = ������−1 Q4. 0.75 Q5. 2.96 × 10−3 ������−1 Q6. I. 0.625 Q8. 0.8 II. 0.75 Q7. 1.01 LEVEL 3 Q8. i) I < II < III < IV Q9. 745.98 ������������ ii) IV < III < II < I TEST PRACTICE PROBLEMS Q. No. Ans. Level Mark (C) / (I) / (U) Q17. C Medium as appropriate Q18. B Medium Q1. D Medium Q19. C Medium Q2. B Medium Q20. A Medium Q3. C Medium Q21. C Medium Q4. A Easy Q22. D Medium Q5. D Medium Q23. B Medium Q6. C Medium Q24. B Hard Q7. A Medium Q25. D Medium Q8. D Medium Q26. A Medium Q9. C Medium Q27. C Easy Q10. A Medium Q28. D Easy Q11. C Hard Q29. C Easy Q12. A Hard Q30. A Easy Q13. A Easy Q31. A Medium Q14. C Hard Q32. B Medium Q15. B Easy Q33. B Hard Q16. C Medium Q34. D Medium
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