Solutions
Maharashtra BoardClass12ChemistryChapter2
Solution
Question 1.
Choose the most correct option.
(i) 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
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
(d) 12 mm Hg
(ii) The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
(c) osmotic pressure
(iii) In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass percent
(a) molarity
(iv) Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1m
(b) 1M
(c) 1 mass%
(d) 1 mole fraction of solute.
(a) 1m
(v) Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
(a) nature of solvent
(vi) Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
(b) boiling point of solvent is lower than that of solution
(vii) 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?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
(c) 4.92 atm
(viii) 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)
(a) 5.41 %
(b) 3.54 %
(c) 4.53 %
(d) 53.4 %
(a) 5.41 %
(ix) Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
(x) Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature.
(a) boiling point of water involved in cooking is increased
(xi) Henry’s law constant for a gas CH_{3}Br is 0.159 moldm3 atm at 250 ^{0}C. What is the solubility of CH_{3}Br in water at 25 ^{0}C and a partial pressure of 0.164 atm?
(a) 0.0159 mol L^{1}
(b) 0.164 mol L^{1}
(c) 0.026 M
(d) 0.042 M
(c) 0.026 M
(xii) Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
(c) they are isotonic solutions
Question 2.
Answer the following in one or two sentences
(i) What is osmotic pressure?
Osmotic pressure : The osmotic pressure is deﬁned as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.
(ii) A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why?
While calculating osmotic pressure by equation, p = CRT, the concentration is expressed in molarity but not in molality.
This is because the measurements of osmotic pressure are made at a certain constant temperature.
Molality depends upon temperature but molality is independent of temperature.
Hence in osmotic pressure measurements, concentration is expressed in molarity.
(iii) Write the equation relating boiling point elevation to the concentration of solution.
The boiling point elevation is directly proportional to the molality of the solution. Thus,
ΔT_{b} ∝ m or ΔT_{b} = K_{b}m
where m is the molality of solution. The proportionality constant K_{b} is called boiling point elevation constant or molal elevation constant or ebullioscopic constant.
If m = 1, ΔT_{b} = K_{b}
Thus, ebullioscopic constant is the boiling point elevation produced by 1 molal solution.
(iv) A 0.1 m solution of K_{2}SO_{4} in water has freezing point of 4.3 ^{0}C. What is the value of van’t Hoff factor if K_{f} for water is 1.86 K kg mol^{1}?
Given : m = 0.1 m, ΔT_{f} = 0  ( 0.43) = 0.43 ^{0}C, K_{f }= 1.86 K kg mol^{1}, i=?
ΔT_{f} = i x K_{f} x m
∴ i = \(\frac{ΔT_f}{k_f×m}=\frac{0.43}{1.86×0.1}\)= 2.312
Answer is van’t Hoff factor = i = 2.312.
(v) What is van’t Hoff factor?
Van’t Hoff factor : To obtain the colligative properties of electrolyte solutions by using relations for nonelectrolytes, van’t Hoff suggested a factor i.
It is deﬁned as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.
The van’t Hoff factor can be represented as,
This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,
\(i=\frac{\text{Observed lowering of vapour pressure}}{\text{Theoretical lowering of vapour pressure}}\)
= ΔP_{(ob)}/ΔP_{(th)}
\(i=\frac{\text{Observed elevation in boiling point}}{\text{Theoretical elevation in boiling point}}\)
= ΔT_{b(ob)}/ΔT_{b(th)}
\(i=\frac{\text{Observed depression in freezing point}}{\text{Theoretical depression in freezing point}}\)
= ΔT_{f(ob)}/ΔT_{f(th)}
\(i=\frac{\text{Observed osmotic pressure}}{\text{Theoretical osmotic pressure}}\)
= π_{(ob)}/ π_{(th)}
 When the solute neither undergoes dissociation or association in the solution, then, i = 1
 When the solute undergoes dissociation in the solution, then, i > l
 When the solute undergoes association in the solution, then i < 1
From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.
van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.
(vi) How is van’t Hoff factor related to degree of ionization?
Consider 1 dm^{3} of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^{y+} ions and y number of B^{x} ions.
Let a be the degree of dissociation.
At equilibrium,
AxBy ⇔ xA^{y+ }+ yB^{x}
For 1 mole of electrolyte : 1 − α, xα, yα
and For ‘m’ moles of an electrolyte : m(1 − α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 − α) + mxα + myα
= m[(1 − α) + xα + yα]
= m[1 + xα + yα − α]
= m[1 + α(x + y − 1)]
The van’t Hoff factor i will be,
\(i=\frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}\)
∴\(i=\frac{m[1 + α(x + y − 1)]}{m}\)
i = 1 + α(x + y − 1)
If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n − 1)
∴α(n − 1) = i − 1
∴ α = \(\frac{i1}{n1}\) …..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.(degree of ionization)
(vii) Which of the following solutions will have higher freezing point depression and why ?
a. 0.1 m NaCl
b. 0.05 m Al_{2}(SO_{4})_{3}
 Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
 More the number of particles in the solution, higher is the depression in freezing point.
The number of particles (ions) from electrolytes are,
a, 0.1 m NaCl
NaCl —> Na^{+} + Cl^{} (Total ions = 0.1 + 0.1=0.2m)
b. 0.05 m Al_{2}(SO_{4})_{3}
Al_{2}(SO_{4})_{3} > 2Al_{3}^{+} + 3SO_{4}^{1 }, (Total ions = 2 x 0.05 + 3 x 0.05 =0.1 +0.15 = 0.25 m)
Therefore, Al_{2}(SO_{4})_{3} solution will have higher freezing point depression. .
(viii) State Raoult’s law for a solution containing a nonvolatile solute
Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.
Explanation : Let P_{0} and P be the vapour pressures of a pure solvent and a solution respectively. If x_{1} is the mole fraction of the solvent then Raoult’s law can be represented as, P = x_{1}P_{0}
For a binary solution containing one solute, if x_{1 }and x_{2} are mole fractions of a solvent and a solute respectively then,
x_{1 }+ x_{2 }= 1
x_{1} = 1  x_{2}
P = x_{1}P_{0}
= (1  x_{2})P_{0}
= P_{0 } x_{2}P_{0}
P_{0}  P = x_{2}P_{0}
∴ x_{2 }= \(\frac{P_0P}{P_0}\)
P_{0}  P = ΔP is the lowering of vapour pressure
∴ x_{2 } = \(\frac{ΔP}{P_0}\)
In the equation, is called relative lowering of vapour pressure.
Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.
(ix) What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm^{3} of water? Why?
 The boiling point of water (or any liquid) depends on its vapour pressure.
 Higher the vapour pressure, lower is the boiling point.
 When 1 mole of volatile methyl alcohol is added to 1 dm^{3} of water, its vapour pressure is increased decreasing the boiling point of water.
(x) Which of the four colligative properties is most often used for molecular mass determination? Why?
 Since osmotic pressure has large values, it can be measured more precisely.
 The osmotic pressure can be measured at a suitable constant temperature.
 The molecular masses can be measured more accurately.
 Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.
Question 3.
Answer the following.
(i) How vapour pressure lowering is related to a rise in boiling point of solution?
 The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 x 10^{3} Nm^{2}). When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
 If the liquid has a low vapour pressure, it has a higher boiling point.
To understand the elevation of boiling point, let us compare the vapour pressures of solution and those of pure solvent. The vapour pressures of solution and of pure solvent are plotted as a function of temperature as shown in Fig.
When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
If T_{0} and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_{b} = T − T_{0}
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.
This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P_{0 }− P, where P_{0} and P are the vapour pressures of the pure solvent and the solution. [ΔT_{b} ∝ (P_{0 }− P) or ΔT_{b} ∝ ΔP]
(ii) What are isotonic and hypertonic solutions?
Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.
Explanation : Consider two solutions of the substances A and B having osmotic pressures π_{A} and π_{B. }If π_{B} is less than π_{A}, then the solution B is a hypotonic solution with respect to the solution A.
Hence, if C_{A} and C_{B} are their concentrations, then, C_{B} < C_{A}. Hence, for equal volumes of the solutions, n_{B} < n_{A}.
Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.
Explanation : Consider two solutions of substances A and B having osmotic pressures π_{A} and π_{B. }If p_{B} is greater than π_{A}, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_{A} and C_{B} are their concentrations, then, C_{B} > C_{A}. Hence, for equal volumes of the solutions, n_{B} > n_{A}.
(iii) A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
 When a solvent and a solution containing a nonvolatile solute are separated by a semipenneable membrane, there arises a ﬂow of solvent molecules from solvent to solution as well as from solution to solvent.
 Due to higher vapour pressure of solvent than solution, the rate of ﬂow of solvent molecules from solvent to solution is higher.
 As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward ﬂow increases.
 At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.
(iv) The osmotic pressure of CaCl_{2} 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 CaCl_{2}
Given : π_{cacl2} = 0.605 atm, π_{urea} = 0.245 atm
For urea solution, van’t Hoff factor,
i = 1
π_{cacl2} = i x (CRT)_{CaCl2}
π_{urea} = (CRT)_{ urea}
π_{cacl2}/π_{urea} = (i x (CRT)_{CaCl2})/(CRT)_{ urea}
∴ i = π_{cacl2}/π_{urea }= 0.605/0.245 = 2.47
Answer is : van’t Hoff factor = i = 2.47
(v) Explain reverse osmosis.
Reverse osmosis : osmosis is a flow of solvent through a semipermeable membrane into the solution. The direction of osmosis can be reversed by applying a pressure larger than the osmotic pressure.
The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like sea water into pure water through a semipermeable membrane is called reverse osmosis.
The osmotic pressure of sea water is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure Water from sea water enters the other side of pure water.
For this purpose a suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.
This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.
(vi) How molar mass of a solute is determined by osmotic pressure measurement?
Consider V dm^{3} (litres) of a solution containing W_{2} mass of a solute of molar mass M_{2} at a temperature T.
Number of moles of solute, n_{2} = W_{2}/M_{2}
The osmotic pressure p is given by,
π = \(\frac{W_2RT}{M_2V}\)
∴ \(M_2=\frac{W_2RT}{πV}\)
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.
Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances
(vii) Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.
 If P_{0} is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P_{0}, then, (P_{0} − P) is the lowering of the vapour pressure.
(viii) Using Raoult’s law, how will you show that ΔP = P^{0}_{1} x_{2} ? Where x_{2} is the mole fraction of solute in the solution and P^{0}_{1} vapour pressure of pure solvent.
If x_{1} and x_{2} are the mole fractions of a solvent and a solute respectively,
then x_{1} + x_{2} = 1
By Raoult’s law, P = x_{1}P_{0. }
where P_{0} is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.
∴ P/P_{0} = x_{1}
1− P/P_{0} = 1 − x_{1}
= x_{2}
If P_{0} − P = ΔP
then ΔP/ P_{0} = x_{2}.
∴ΔP = P_{0}x_{2}
(ix) While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
 Boiling point elevation and freezing point depression involve temperature changes, (ΔT_{b} and ΔT_{f}).
 Since molarity depends on temperature but molality is independent of temperature we use molality and not molarity in considering boiling point elevation and freezing point depression.
Question 4.
Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
Consider 1 dm^{3} of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^{y+} ions and y number of B^{x} ions.
Let a be the degree of dissociation.
At equilibrium,
AxBy ⇔ xA^{y+ }+ yB^{x}
For 1 mole of electrolyte : 1 − α, xα, yα
and For ‘m’ moles of an electrolyte : m(1 − α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 − α) + mxα + myα
= m[(1 — α) + xα + yα]
= m[1 + xα + yα − α]
= m[1 + α(x + y − 1)]
The van’t Hoff factor i will be,
\(i=\frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}\)
∴\(i=\frac{m[1 + α(x + y − 1)]}{m}\)
i = 1 + α(x + y − 1)
If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n − 1)
∴α(n − 1) = i −1
∴ α = \(\frac{i−1}{n−1}\) …..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.
Question 5.
What is effect of temperature on solubility of solids in water? Give examples.
The solubility of a solid solute depends upon temperature.
Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.
Figure shows the result of experimental determination of solubilities of some ionic solids in water at various temperatures.
Following are some experimental observations from Fig
 Dissolution process may be endothermic Or exothermic.
 Solubilities of NaBr, NaCl and KCl change slightly with temperature.
 Solubilities of KNO_{3}, NaNO_{3} and KBr increase appreciably with increasing temperature.
 Solubility of Na_{2}SO_{4} decreases with increase of temperature.
This variation in solubility with temperature can be used to separate the salts from the mixture by fractional clystallisation.
Question 6.
Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelctrolyte and its molar mass.
The freezing point depression, ΔT_{f} of a solution is directly proportional to molality (m) of the solution.
ΔT_{f} ∝ m
ΔT_{f} = K_{f }m
where Kf is a molal depression constant.
The molality of a solution is given by,
m = \(\frac{\text{ Number of moles of the solute}}{\text{Weight of the solvent in kg}}\)
If W_{1} grams of a solvent contain W_{2} grams of a solute of the molar mass M_{2}, then the molality m of the solution is given by,
m = \(\frac{W_2×1000}{W_1M_2}=\frac{W_2}{W_1M_2}\) mol kg^{−1}
∴ ΔT_{f} = K_{f} x \(\frac{W_2×1000}{W_1M_2}\)
∴ M_{2 }= \(\frac{K_fW_2×1000}{ΔT_fW_1M_2}\) or M_{2 }= \(\frac{K_fW_2}{ΔT_fW_1M_2}\) mol kg^{−1}
If the weights and molecular weight are expressed in kg, then,
∴ ΔT_{f} = K_{f} x \(\frac{W_2}{W_1M_2}\)
The unit of K_{f} is K kg mol^{1} .
Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.
Question 7.
Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
 The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 x 10^{3} Nm^{2}). When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
 If the liquid has a low vapour pressure, it has a higher boiling point.
To understand the elevation of boiling point, let us compare the vapour pressures of solution and those of pure solvent. The vapour pressures of solution and of pure solvent are plotted as a function of temperature as shown in Fig.
When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
If T_{0} and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_{b} = T − T_{0}
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.
This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P_{0 }− P, where P_{0} and P are the vapour pressures of the pure solvent and the solution. [ΔT_{b} ∝ (P_{0 }− P) or ΔT_{b} ∝ ΔP]
Question 8.
Fish generally needs O_{2} concentration in water at least 3.8 mg/L for survival. What partial pressure of O_{2} above the water is needed for the survival of fish? Given the solubility of O_{2} in water at 0 ^{0}C and 1 atm partial pressure is 2.2 × 10^{3} mol/L
Given : Required concentration of O_{2} = 3.8 mg/L =
Solubility of O_{2} = 2.2 x 10^{3} mol L^{1}
P = 1 atm
Partial pressure of O_{2} needed = Po_{2} = ?
S = K_{H} x Po_{2 }
∴ K_{H }= S/ Po_{2} = (2.2 x 10^{3})/1 mol L^{1} atm^{1}
= 2.2 x 10^{3} mol L^{1} atm^{1}
Po_{2 }= S/K_{H} =
= 0.05397 atm
Ans. Pressure needed = Po_{2} = 0.05397 aim.
Question 9.
The vapour pressure of water at 20^{0}C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water?
Given : Vapour pressure of pure solvent (water) = P_{0} = 17 mm Hg
Weight of solvent = W_{1} = 50 g
Weight of solute (urea) = W_{2} = 2.8 g
Molecular weight of a solvent = M_{1} = 18
Molecular weight of a solute (urea) = M_{2 }= 60 g mol^{1}
\(\frac{P_0P}{P_0} =\frac{W_2/M_2}{W_1/M_1}\)
\(\frac{17P}{17} =\frac{2.8×18}{50×60}=0.0168\)
17P = 17 x 0.0168
∴ P = 17 − 0.2856 = 16.7144 mm Hg
Vapour pressure of solution = 16.7144 mm Hg
Question 10.
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution.
Given : W_{2} = 5 g cane sugar, W_{1} = 100 − 5 = 95 g, M_{2} = 342 g mol^{1}, T_{f1} = 271 K;
ΔT_{f1} =273 − 271 = 21K, T_{f} = ?
W_{2}’ = 5 g glucose, W_{1}’ = 100 − 5 = 95 g,
M_{2}’ = 180 g mol^{1}, ΔT_{f2} = ?
ΔT_{f} = K_{f} x \(\frac{W_2×1000}{W_1M_2}\)
K_{f }= \(\frac{ΔT_fW_1M_2}{W_2×1000}\)
ΔT_{f2} = K_{f} x \(\frac{W'_2×1000}{W'_1M'_2}=\frac{13×5×1000}{95×180}=3.801 K\)
Freezing point of solution = T_{f} = = 273 − 3.801 = 269.199 K
Freezing point of solution = 269.2 K.
Question 11.
A solution of citric acid C_{6}H_{8}O_{7} in 50 g of acetic acid has a boiling point elevation of 1.76 K. If K_{b} for acetic acid is 3.07 K kg mol^{1}, what is the molality of solution?
Given : W_{1} = 50 g acetic acid
ΔT_{b} = 1.76 K
K_{b} = 3.07 K kg mol^{1}
m = ?
ΔT_{b} = K_{b} x m
m = ΔT_{b}/ K_{b }= 1.76/3.07
= 0.5733 m
Molality of solution = 0.5733 m
Question 12.
An aqueous solution of a certain organic compound has a density of 1.063 gmL^{1}, an osmotic pressure of 12.16 atm at 25^{0}C and a freezing point of −1.03 ^{0}C. What is the molar mass of the compound?
Given: Density of a solution d = 1.063 g mL^{1 }= 1.063, Osmotic pressure of solution π = 12.16 atm, Temperature T = 25 °C = 298.15 K,
Freezing point of solution = T_{f} = –1.03 ^{0}C, ∴ ΔT_{f} = 0 – (−1.03) =1.03 ^{0}C
R = 0.08206 dm^{3} atm K^{1} mol^{1}, K_{f} of water = 1.86 K kg mol^{1}
ΔT_{f} = K_{f }m
∴ m = ΔT_{f}/ K_{f} = 1.03/ 1.86 = 0.554 mol kg^{1} = 0.554 M
π = CRT
12.16 = C x 0.08206 x 298.15
∴ C = 0.497 mol dm^{–3} = 0.497 M
Mass of solvent W_{1 }= C/m = (0.497/0.554) x 1 dm^{3} = 0.897 kg = 897 g
Mass of Solution = 1.063 x g mL^{–1 }x 1000 mL = 1063g = 1.063 kg =1063 g
∴ Mass of Solute W_{2 }= 1.063 – 0.897 = 0.166 kg =166 g
m = \(\frac{W_2×1000}{W_1M_2}\)
∴ M_{2} = \(\frac{W_2×1000}{W_1m}= \frac{166g×1000g\,Kg^{1}}{897g×0.554mol\,kg^{1}}\) = 334 g mol^{1}
Molar mass of the compound = 334 g mol^{1}
Question 13.
A mixture of benzene and toluene contains 30% by mass of toluene. At 30^{0}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^{0}C.
Given : 30% by mass of toluene (T) and 70% by mass of benzene (B).
W_{T} = 30 g, W_{B} = 70 g, P^{0}_{T} = 36.7 mm Hg, P^{0}_{B} = 118.2 mmHg
M_{T }= 92 g mol^{1}, M_{B} = 78 g mol^{1}
P_{T} = ?, P_{B} = ?, P_{soln} = ?
n_{T} = W_{T}/ M_{T} = 30/92 = 0.3260 mol
n_{B} = W_{B}/ M_{B} = 70/78 = 0.8974 mol
Total number of moles = n_{Total} = n_{T} + n_{B}
= 0.326 + 0.8974
= 1.2234 mol
Mole fractions :
x_{T} = n_{T}/ n_{Total} = 0.326/1.2234 = 0.2665
x_{B} = 10.2665 = 0.7335
P_{soln} = x_{T} x P^{0}_{T} + x_{B} x P^{0}_{B}
= 0.2665 x 36.7 + 0.7335 x 118.2
= 9.780 + 86.7
= 96.48 mm Hg
Partial pressures :
P_{T} = x_{T} x P_{soln}
= 0.2665 x 96.48
= 25.71 mm Hg
P_{B} = x_{B} X P_{soln}
= 0.7335 x 96.48
= 70.77 mm Hg
Answer is : Total pressure = P_{soln} = 96.48 mm Hg
Partial pressures Toluene : P_{T} = 25.71 mm Hg
Partial pressures Benzene : P_{B} = 70.77 mm Hg
Question 14.
At 25 ^{0}C a 0.1 molal solution of CH_{3}COOH is 1.35 % dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.
Given: T= 273 + 25 = 298 K, C= 0.1 m ≅ 0.1 M, K_{f }= 1.86 K kg mol^{1},
Per cent dissociation = 1.35, Freezing point = t_{f} = ?, p = ?
α = 1.35/100 = 0.013
CH_{3}COOH ↔ CH_{3}COO + H^{+}
1 − α α α
i = 1 − α + α + α = 1 + α = 1 + 0.0135 = 1.0135
(i) ΔT_{f }= i x K_{f} x m
= 1.0135 x 1.86 x 0.1
= 0.1885 °C
Freezing point of solution = 0 − 0.1885 = − 0.1885 °C
(ii) π = iCRT
= 1.035 x 0.1 x 0.08206 x 298
= 2.53 atm
Answer is : (i) Freezing point of solution = − 0.1885 °C
(ii) Osmotic pressure = π = 2.53 atm.
Question 15.
A 0.15 m aqueous solution of KCl freezes at −0.510 ^{0}C. Calculate i and osmotic pressure at 0 ^{0}C. Assume volume of solution equal to that of water
Given : c = 0.15 m KCl ≅ 0.15 M KCl, ΔT_{f }= 0 − T_{f }= 0 − (− 0.510) = 0.510 °C
T = 273 K, K_{f }= 1.86 K kg mol^{1}
i = ?; π = ?
ΔT_{f }= i x K_{f} x m
∴ i = \(\frac{ΔT_f}{K_fm}\)
π = iCRT
= 1.828 x 0.15 x 0.08206 x 273
= 6.143 atm
Answer is i = 1.828, Osmotic pressure = p = 6.143 atm
Important Formulae :
Henry’s law : S = K_{H}P Raoult’s law : P_{soln} = x_{1}P_{0} Mole fractions, x_{1} + x_{2} = 1 P_{soln} = _{ }P^{0}_{1} x_{1} + P^{0}_{2} x_{2} P_{soln} = (P^{0}_{2}  P^{0}_{1})x_{2 }+ P^{0}_{1} Mole fractions of components in vapour phase : and For ideal solutions : ΔV_{mix} = 0; Δ_{mix} H = 0 Relative lowering of vapour pressure = \(\frac{p^0P}{P^0}\), \(x_2=\frac{p^0P}{P^0}\) \(\frac{p^0P}{P^0}=\frac{W_2×M_1}{W_1×M_2}\) ΔT_{b }= K_{b} x m and ΔT_{f }= K_{f} x m ΔT_{b }= K_{b} x \(\frac{W_2×1000}{W_1×M_2}\) ΔT_{f }= K_{f} x \(\frac{W_2×1000}{W_1×M_2}\) π = cRT and π = \(\frac{WRT}{MV}\) \(i=\frac{\text{Colligative property of electrolyte solution}}{\text{Colligative property of nonelectrolyte solution of the same concentration}}\) van’t Hoff factor (i) = \(\frac{ΔT_{b(ob)}}{ΔT_{b(th)}}\) = \(\frac{ΔP_{(ob)}}{ΔP_{(th)}}\) = \(\frac{ΔT_{f(ob)}}{ΔT_{f(th)}}\) = \(\frac{π_{(ob)}}{π_{(th)}}\) = \(\frac{M_{(ob)}}{M_{(th)}}\) Colligative properties considering van’t Hoff factor :

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