Page 1 :
Solutions, EXERCISES [PAGES 44 - 46], Exercises | Q 1.01 | Page 44, Choose the most correct option., The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water, (vapour pressure of pure water = 24 mm Hg) is ______., 1. 24 mm Hg, 2. 32 mm Hg, 3. 48 mm Hg, 4. 12 mm Hg, Solution: The vapour pressure of a solution containing 2 moles of a solute in 2 moles of, water (vapour pressure of pure water = 24 mm Hg) is 12 mm Hg., Exercises | Q 1.02 | Page 44, Choose the most correct option., The colligative property of a solution is _______., 1. vapour pressure, 2. boiling point, 3. osmotic pressure, 4. freezing point, Solution: The colligative property of a solution is osmotic pressure., Exercises | Q 1.03 | Page 44, Choose the most correct option., In calculating osmotic pressure the concentration of solute is expressed in _______., 1. molarity, 2. molality, 3. mole fraction, 4. mass percent, Solution: In calculating osmotic pressure the concentration of solute is expressed, in molarity., Exercises | Q 1.04 | Page 44, Choose the most correct option.

Page 2 :
Ebullioscopic constant is the boiling point elevation when the concentration of a solution, is _______., 1. 1 m, 2. 1 M, 3. 1 mass %, 4. 1-mole fraction of solute., Solution: Ebullioscopic constant is the boiling point elevation when the concentration of, solution is 1 m., Exercises | Q 1.05 | Page 44, Choose the most correct option., Cryoscopic constant depends on _______., 1. nature of solvent, 2. nature of solute, 3. nature of solution, 4. number of solvent molecules, Solution: Cryoscopic constant depends on number of solvent molecules., Exercises | Q 1.06 | Page 45, Choose the most correct option., Identify the CORRECT statement., 1. Vapour pressure of solution is higher than that of pure solvent., 2. Boiling point of solvent is lower than that of solution., 3. Osmotic pressure of solution is lower than that of solvent., 4. Osmosis is a colligative property., Solution: Boiling point of solvent is lower than that of solution., Exercises | Q 1.07 | Page 45, Choose the most correct option., A living cell contains a solution which is isotonic with 0.3 M sugar solution. What, osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body, temperature?, 1. 5.08 atm, 2. 2.54 atm, 3. 4.92 atm

Page 3 :
4. 2.46 atm, Solution: 2.54 atm., Exercises | Q 1.08 | Page 45, Choose the most correct option., The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose, isotonic with blood has the percentage (by volume)________., 1. 5.41 %, 2. 3.54 %, 3. 4.53 %, 4. 53.4 %, Solution: The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of, glucose isotonic with blood has the percentage (by volume) 5.41%., , Exercises | Q 1.09 | Page 45, Choose the most correct option., Vapour pressure of a solution is _______., 1. directly proportional to the mole fraction of the solute, 2. inversely proportional to the mole fraction of the solute, 3. inversely proportional to the mole fraction of the solvent, 4. directly proportional to the mole fraction of the solvent, Solution:, Vapour pressure of a solution is inversely proportional to the mole fraction of the, solute., Exercises | Q 1.1 | Page 45

Page 4 :
Choose the most correct option., Pressure cooker reduces cooking time for food because _______., 1. boiling point of water involved in cooking is increased, 2. heat is more evenly distributed in the cooking space, 3. the higher pressure inside the cooker crushes the food material, 4. cooking involves chemical changes helped by a rise temperature, Solution:, Pressure cooker reduces cooking time for food because boiling point of water, involved in cooking is increased., Exercises | Q 1.11 | Page 45, Choose the most correct option., Henry’s law constant for a gas CH3Br is 0.159 mol dm-3 atm-1 at 25 °C. What is the, solubility of CH3Br in water at 25 °C and partial pressure of 0.164 atm?, 1. 0.0159 mol L-1, 2. 0.164 mol L-1, 3. 0.026 M, 4. 0.042 M, Solution: 0.026 M, Explanation:, S = KHP = 0.159 mol dm-3 atm-1 × 0.164 atm, = 0.026 M, Exercises | Q 1.12 | Page 45, Choose the most correct option., Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M, sucrose solution?, 1. Osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose, solution, 2. Urea solution is hypertonic to sucrose solution, 3. They are isotonic solutions, 4. Sucrose solution is hypotonic to urea solution, Solution: They are isotonic solutions.

Page 5 :
Exercises | Q 2.01 | Page 45, Answer the following in one or two sentences., What is osmotic pressure?, Solution:, The hydrostatic pressure (on the side of solution) that stops osmosis is called an, osmotic pressure of the solution., OR, The excess of pressure on the side of the solution that stops the net flow of solvent into, the solution through a semipermeable membrane is called osmotic pressure., Exercises | Q 2.02 | Page 45, Answer the following in one or two sentences., A solution concentration is expressed in molarity and not in molality while considering, osmotic pressure. Why?, Solution:, 1. The osmotic pressure measurements are made at a specific constant temperature., Molarity remains constant at a specific temperature., 2. It is not necessary to express concentration in a temperature-independent unit like, molality., Hence, the solute concentration is expressed in molarity while calculating osmotic, pressure rather than molality., Exercises | Q 2.03 | Page 45, Answer the following in one or two sentences., Write the equation relating boiling point elevation to the concentration of the solution., Solution:, The boiling point elevation is directly proportional to the molality of the solution. Thus,, Δ Tb ∝ m, ∴ Δ Tb ∝ Kb m, where, m is the molality of solution. The proportionality constant Kb is called boiling, point elevation constant or molal elevation constant or ebullioscopic constant.

Page 6 :
Exercises | Q 2.04 | Page 45, Answer the following in one or two sentences., A 0.1 m solution of K2SO4 in water has a freezing point of – 4.3 °C. What is the value of, van’t Hoff factor if Kf for water is 1.86 K kg mol–1?, Solution:, Given: Molality of K2SO4 solution = m = 0.1 m, Freezing point of solution = Tf = – 4.3 °C, Kf of water = 1.86 K kg mol–1, To find: van’t Hoff factor, Formula: ΔTf = i Kf m, , Exercises | Q 2.05 | Page 45, Answer the following in one or two sentences., What is van’t Hoff factor?, Solution:, van’t Hoff factor (i) is defined as the ratio of colligative property of a solution of, electrolyte divided by the colligative property of nonelectrolyte solution of the same, concentration., Thus,

Page 7 :
where quantities without subscript refer to electrolytes and those with subscript to, nonelectrolytes., Exercises | Q 2.06 | Page 45, Answer the following in one or two sentences., How is van’t Hoff factor related to degree of ionization?, Solution:, The van’t Hoff factor is related to degree of ionization as follows:, i = 1 + α (n – 1), or, α = i-1/n-1, where, α = Degree of ionization/dissociation, i = van’t Hoff factor, n = Moles of ions obtained from ionization of 1 mole of electrolyte., Exercises | Q 2.07 | Page 45, Answer the following in one or two sentences., Which of the following solution will have higher freezing point depression and why?, i. 0.1 m NaCl, ii. 0.05 m Al2(SO4)3, Solution:, For 0.1 m NaCl:, NaCl → Na+, 0.1 m, , 0.1 m, , Cl-, , +, , 0.1 m, , Total particles in solution = 0.2 mol, For 0.05 m Al2(SO4)3:

Page 8 :
0.05 m, , 0.1 m, , 0.15 m, , Total particles in solution = 0.25 mol, Al2(SO4)3 solution contains more number of particles than NaCl solution. Hence,, Al2(SO4)3 solution has maximum ΔTf., Therefore, the freezing point depression of 0.05 m Al2(SO4)3 solution will be higher than, 0.1 m NaCl solution., Exercises | Q 2.08 | Page 45, Answer the following in one or two sentences., State Raoult’s law for a solution containing a nonvolatile solute., Solution:, The Raoult’s law states that, “the vapour pressure of solvent over the solution is equal, to the vapour pressure of pure solvent multiplied by its mole fraction in the solution.”, Exercises | Q 2.09 | Page 45, Answer the following in one or two sentences., What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1, dm3 of water? Why?, Solution:, i. When 1 mole of methyl alcohol is added to 1 dm3 of water, the boiling point of water, decreases., ii. Methyl alcohol is a volatile liquid. Therefore, it increases the vapour pressure of a, solution at a given temperature. Hence, the solution boils at lower temperature., Exercises | Q 2.1 | Page 45, Answer the following in one or two sentences., Which of the four colligative properties is most often used for molecular mass, determination? Why?, Solution:, i. Among the four colligative properties, osmotic pressure is most often used for, molecular mass determination., ii. Osmotic pressure is much larger and therefore more precisely measurable property, than other colligative properties.

Page 9 :
Therefore, it is useful to determine molar masses of very expensive substances and of, the substances that can be prepared in small quantities., Exercises | Q 3.1 | Page 45, Answer the following., How vapour pressure lowering is related to a rise in the boiling point of solution?, Solution:, i. At the boiling point of a liquid, its vapour pressure is equal to 1 atm., ii. In order to reach boiling point, the solution and solvent must be heated to a, temperature at which their respective vapour pressures attain 1 atm., iii. At any given temperature the vapour pressure of a solution is lower than that of pure, solvent. Hence, the vapour pressure of solution needs a higher temperature to reach 1, atm than that of needed for vapour pressure of solvent., Therefore, vapour pressure lowering causes a rise in the boiling point of a solution., Exercises | Q 3.2 | Page 45, Answer the following., What are isotonic and hypertonic solutions?, Solution:, i. Isotonic solutions:, Two or more solutions having the same osmotic pressure are said to be isotonic, solutions., e.g. For example, 0.1 M urea solution and 0.1 M sucrose solution are isotonic because, their osmotic pressures are equal. Such solutions have the same molar concentrations, but different concentrations in g/L. If these solutions are separated by a semipermeable, membrane, there is no flow of solvent in either direction., ii. Hypertonic solution:, If two solutions have unequal osmotic pressures, the more concentrated solution with, higher osmotic pressure is said to be the hypertonic solution., e.g. For example, if osmotic pressure of sucrose solution is higher than that of urea, solution, the sucrose solution is hypertonic to urea solution., Exercises | Q 3.3 | Page 46, Answer the following.

Page 10 :
A solvent and its solution containing a nonvolatile solute are separated by a, semipermeable membrane. Does the flow of solvent occur in both directions? Comment, giving a reason., Solution:, 1. When a solution and pure solvent or two solutions of different concentrations are, separated by a semipermeable membrane, the solvent molecules pass through the, membrane., 2. The passage of solvent molecules through the semipermeable membrane takes, place in both directions, since the solvent is on both sides of the membrane., 3. However, the rate of passage of solvent molecules into the solution or from a more, dilute solution to more concentrated solution is found to be greater than the rate in the, reverse direction., 4. This is favorable since the vapour pressure of solvent is greater than that of solution., Exercises | Q 3.4 | Page 46, Answer the following., The osmotic pressure of CaCl2 and urea solutions of the same concentration at the, same temperature are respectively 0.605 atm and 0.245 atm, calculate van’t Hoff factor, for CaCl2., Solution:, Given: Osmotic pressure of CaCl2 solution = 0.605 atm, Osmotic pressure of urea solution = 0.245 atm, To find: The value of van’t Hoff factor, Formulae: π = MRT, π = iMRT, Calculation: For urea solution, π = MRT, 0.245 atm = MRT, , ....(i), , For CaCl2 solution, π = iMRT, 0.602 atm = iMRT, , ....(ii), , From equations (i) and (ii),

Page 11 :
∴ i = 2.47, The value of van’t Hoff factor is 2.47., Exercises | Q 3.5 | Page 46, Answer the following., Explain reverse osmosis., Solution:, i. If a pressure larger than the osmotic pressure is applied to the solution side, then pure, solvent from the solution passes into pure solvent side through the semipermeable, membrane. This phenomenon is called reverse osmosis., ii. For example, consider fresh water salt water separated by a semipermeable, membrane. When the pressure larger than the osmotic pressure of a solution is applied, to solution, pure water from salty water passes into fresh pure water through the, membrane. Thus, the direction of osmosis can be reversed by applying a pressure, larger than the osmotic pressure., iii. The schematic set up for reverse osmosis is as follows:, , Exercises | Q 3.6 | Page 46, Answer the following., How molar mass of a solute is determined by osmotic pressure measurement?, Solution:, 1. For very dilute solutions, the osmotic pressure follows the equation,

Page 12 :
This formula can be used for the calculation of molar mass of a nonionic solute (i.e.,, nonelectrolyte), by osmotic pressure measurement., Exercises | Q 3.7 | Page 46, Answer the following., Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?, Solution:, i. Vapour pressure of a liquid depends on the ease with which the molecules escape, from the surface of the liquid., ii. When a nonvolatile solute is dissolved in a solvent, some of the surface molecules of, the solvent are replaced by nonvolatile solute molecules. These solute molecules do not, contribute to vapour above the solution., iii. Thus, the number of solvent molecules available for vaporization per unit surface, area of a solution is less than the number at the surface of the pure solvent., iv. As a result the solvent molecules at the surface of solution vaporize at a slower rate, than pure solvent. This results in lowering of vapour pressure, Exercises | Q 3.8 | Page 46, Answer the following., , Solution:

Page 13 :
1. Raoult’s law expresses the quantitative relationship between vapour pressure of, solution and vapour pressure of the solvent., 2. In solutions of nonvolatile solutes, the law is applicable only to the volatile solvent., 3. The law states that, “the vapour pressure of solvent over the solution is equal to the, vapour pressure of pure solvent multiplied by its mole fraction in the solution.”, 4. Suppose that for a binary solution containing solvent and one nonvolatile solute, P 1 is, the vapour pressure of, solvent over the solution, x1 and x2 are the mole fractions of solvent and solute,, respectively and, , is the vapour pressure of pure solvent, then,, , 5. Since, x1 = 1 – x2,, , Note: A plot of P1 versus x1 is a straight as shown below., Variation of vapour pressure of solution with mole fraction of solvent, , Exercises | Q 3.9 | Page 46, Answer the following., While considering boiling point elevation and freezing point depression a solution, concentration is expressed in molality and not in molarity. Why?, Solution:, 1. In boiling point elevation and freezing point depression, we deal with the systems, whose temperature is not constant.

Page 14 :
2. We cannot express the concentration of the solution in molarity because it changes, with temperature whereas molality is temperature independent., Hence, while considering boiling point elevation and freezing point depression a solution, concentration is expressed in molality and not in molarity., Exercises | Q 4 | Page 46, Derive the relationship between the degree of dissociation of an electrolyte and van’t, Hoff factor., Solution:, 1. The weak electrolytes involve the concept of degree of dissociation (α) that changes, the van’t Hoff factor., 2. Consider an electrolyte AxBy that dissociates in aqueous solution as, AxBy, , ⇌, , Initially, , 1 mol, , At equilibrium, , (1 - α) mol, , xAy+, , +, , 0, (x α mol), , yBx0, (yα mol), , 3. If α is the degree of dissociation of electrolyte, then the moles of cations are xα and, those of anions are yα equilibrium. We have dissolved just 1 mol of electrolyte initially. α, mol of electrolyte dissociates and (1 – α) mol remains undissociated at equilibrium., Total moles after dissociation = (1 – α) + (xα) + (yα), = 1 + α (x + y - 1), = 1 + α (n - 1), where, n = x + y = moles of ions obtained from dissociation of 1 mole of electrolyte., 4. The van’t Hoff factor given as, , Exercises | Q 5 | Page 46, What is the effect of temperature on solubility of solids in water? Give examples., Solution:

Page 15 :
1. The effect of temperature on solubility of a substance depends on enthalpy of, solution., a. When the substance dissolves in water by an endothermic process, that is, with the, absorption of heat, its solubility increases with an increase of temperature., e.g. KCl dissolve in water by endothermic process., b. On the other hand, when the substance dissolves in water by an exothermic process,, that is, with the release of heat, its solubility decreases with an increase of temperature., e.g. CaCl2 and Li2SO4.H2O dissolve in water releasing heat., 2. It is important to understand that there is no direct correlation between solubility and, exothermicity or endothermicity. For example, dissolution of CaCl2 in water is, exothermic and that of NH4NO3 is endothermic. However, the solubility of these, substances increases with the temperature, Exercises | Q 6 | Page 46, Obtain the relationship between freezing point depression of a solution containing, nonvolatile-nonelectrolyte solute and its molar mass., Solution:, 1. The freezing point depression (ΔTf) is directly proportional to the molality of solution., Thus, ΔTf = Kf m, , ….(1), , 2. Suppose we prepare a solution by dissolving W2 g of solute in W1 g of solvent., , 4. Substituting equation (2) in equation (1), we get,

Page 16 :
Exercises | Q 7 | Page 46, Explain with diagram the boiling point elevation in terms of vapour pressure lowering., Solution:, 1. The vapour pressures of a solution and of pure solvent are plotted as a function of, temperature in the given diagram., 2. At any temperature, the vapour pressure of a solution is lower than that of pure, solvent. Hence, the vapour pressure-temperature curve of solution (CD) lies below that, of solvent (AB)., 3. The difference between the two vapour pressures increases as temperature and, vapour pressure increase as predicted by the equation,, , 4. The intersection of the curve CD with the line corresponding to 760 mm is the boiling, point of solution. The similar intersection of the curve AB is the boiling point of pure, solvent. It is clear from the diagram that the boiling point of the solution (Tb) is higher, than that of pure solvent, 5. At the boiling point of a liquid, its vapour pressure is equal to 1 atm., 6. In order to reach boiling point, the solution and solvent must be heated to a, temperature at which their respective vapour pressures attain 1 atm., 7. At any given temperature the vapour pressure of solution is lower than that of pure, solvent. Hence, the vapour pressure of solution needs a higher temperature to reach 1, atm than that needed for vapour pressure of solvent., 8. In other words, the solution must be heated to higher temperature to cause it to boil, than the pure solvent.

Page 17 :
Vapour pressure-temperature, of pure solvent and solution, , Exercises | Q 8 | Page 46, Fish generally needs O2 concentration in water at least 3.8 mg/L for survival. What, partial pressure of O2 above the water is needed for the survival of fish? Given the, solubility of O2 in water at 0 °C and 1 atm partial pressure is 2.2 × 10-3 mol/L., Solution:, Given: O2 concentration in water required for fishes = 3.8 mg/L, Solubility of O2 in water = 2.2 × 10-3 mol/L, Pressure = 1 atm, To find: Partial pressure of O2 above the water needed for the survival of fish., Formula: S = KHP, Calculation: Pressure = 1 atm = 1.013 bar, Now, using formula and rearranging,

Page 18 :
The partial pressure of O2 above the water needed for the survival of fish is 0.0548 bar., Exercises | Q 9 | Page 46, The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of, solution containing 2.8 g urea in 50 g of water?, Solution:, , ∴ 17 mm Hg = 0.0168 × 17 mm Hg, ∴ 17 mm Hg - P1 = 0.2856 mm Hg, ∴ P1 = 17 mm Hg - 0.2856 mm Hg = 16.71 mm Hg, Vapour pressure of the given solution is 16.71 mm Hg., Exercises | Q 10 | Page 46

Page 19 :
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has a freezing, point of 271 K. Calculate the freezing point of 5% aqueous glucose solution., Solution:

Page 20 :
Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W 2), = 5 g, and mass of solvent (W1) = 95 g., Now, using formula (i),

Page 21 :
Exercises | Q 11 | Page 46, A solution of citric acid C6H8O7 in 50 g of acetic acid has a boiling point elevation of 1.76, K. If Kb for acetic acid is 3.07 K kg mol-1, what is the molality of solution?, Solution:

Page 22 :
Exercises | Q 12 | Page 46, An aqueous solution of a certain organic compound has a density of 1.063 g mL -1 ,, osmotic pressure of 12.16 atm at 25 °C and a freezing point of −1.03 °C. What is the, molar mass of the compound?, Solution:, Given: Density of a solution = d = 1.063 g mL-1, Osmotic pressure of solution = π = 12.16 atm, Temperature = T = 25 °C = 298.15 K, Freezing point of solution = Tf = - 1.03 °C, To find: Molar mass of a compound, , Using formula (ii),

Page 23 :
π = MRT, , Exercises | Q 13 | Page 46, A mixture of benzene and toluene contains 30% by mass of toluene. At 30 °C, vapour, pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg., Assuming that the two liquids form ideal solutions, calculate the total pressure and, partial pressure of each constituent above the solution at 30 °C., Solution:

Page 24 :
∴ Partial pressures of toluene and benzene are 9.78 mm Hg and 86.7 mm Hg,, respectively., ∴ Total pressure above the solution is 96.5 mm Hg., Exercises | Q 14 | Page 46

Page 25 :
At 25 °C, a 0.1 molal solution of CH3COOH is 1.35 % dissociated in an aqueous, solution. Calculate the freezing point and osmotic pressure of the solution assuming, molality and molarity to be identical., Solution:, Given:, Molality of solution (m) = 0.1 m = 0.1 mol kg-1, Degree of dissociation (α) = 1.35% = 0.0135, Temperature = 25 °C = 25 °C + 273.15 = 298.15 K, Molarity of solution (M) = 0.1 M, To find:, 1. Freezing point of solution, 2. Osmotic pressure of solution, Formulae:, , Δ Tf = i Kfm = 1.0135 × 1.86 K kg mol-1 × 0.1 mol kg-1, = 0.189 K = 0.189 °C, , π = i MRT = 1.0135 × 0.1 mol dm-3 × 0.08205 dm3 atm K-1 mol-1 × 298.15 K = 2.48 atm

Page 26 :
∴ The freezing point of the solution is – 0.189 °C, ∴ The osmotic pressure of solution at 25 °C is 2.48 atm, Exercises | Q 15 | Page 46, A 0.15 m aqueous solution of KCl freezes at - 0.510 °C. Calculate i and osmotic, pressure at 0 °C. Assume the volume of solution equal to that of water., Solution:, Given:, Molality of solution = m = 0.15 m, Freezing point of solution = Tf = - 0.510 °C, Temperature = 0 °C = 273 K, To find:, 1. The value of van’t Hoff factor (i), 2. Osmotic pressure of solution, Formulae:

Page 27 :
π = 1.83 × 3.36 atm, π = 6.15 atm, ∴ The van’t Hoff factor is 1.83., ∴ The osmotic pressure of solution at 0 °C is 6.15 atm.