Choose the correct statements from the following: (A) two different s…

Choose the correct statements from the following:

(A) two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point.
(B) the osmotic pressure of a solution is given by the equation Π = CRT (where C is the molarity of the solution)
(C) decreasing order of osmotic pressure for 0.01 M aqueous solution of \( BaCl_2, KCl, CH_3COOH \) and sucrose is \( BaCl_2> KCl > CH_3COOH \) > sucrose
(D) according to Raoult's law the vapor pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 3

The correct answer is Option (3) → (B), (C) and (D) only

Colligative properties depend on the number of solute particles, not on their nature. For a given solvent the depression in freezing point is \( \Delta T_f = K_f\,m \), and the osmotic pressure of a dilute solution is \( \Pi = CRT \). For an electrolyte both are multiplied by the van’t Hoff factor i, the number of particles each formula unit gives in solution.

(A) two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point. — Incorrect. In \( \Delta T_f = K_f\,m \), the cryoscopic constant \( K_f \) is a property of the solvent (1.86 K kg mol⁻¹ for water, 5.12 K kg mol⁻¹ for benzene). The same molality in different solvents therefore gives different depressions.
(B) the osmotic pressure of a solution is given by the equation Π = CRT (where C is the molarity of the solution) — Correct. This is the van’t Hoff equation for a dilute solution: \( \Pi = \frac{n_2}{V}RT = CRT \).
(C) decreasing order of osmotic pressure for 0.01 M aqueous solution of \(BaCl_2, KCl, CH_3COOH\) and sucrose is \(BaCl_2> KCl > CH_3COOH\) > sucrose — Correct. At the same C and T, \( \Pi = iCRT \) is proportional to i:
    \( \displaystyle \ce{BaCl2 -> Ba^2+ + 2Cl-}:\ 1 + 2 = 3 \text{ ions},\ i = 3 \)
    \( \displaystyle \ce{KCl -> K+ + Cl-}:\ 2 \text{ ions},\ i = 2 \)
    \( \displaystyle \ce{CH3COOH <=> CH3COO- + H+}:\ \text{weak acid, only slightly ionised},\ i = 1 + \alpha \text{ (just above 1)} \)
    \( \displaystyle \text{Sucrose}:\ \text{does not dissociate},\ i = 1 \)
So \( \ce{BaCl2} \gt \ce{KCl} \gt \ce{CH3COOH} \gt \) sucrose.
(D) according to Raoult’s law the vapor pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution — Correct. Raoult’s law states \( p_1 = p_1^{0}\,x_1 \): the partial vapour pressure of each volatile component is proportional to its mole fraction, the constant being its vapour pressure in the pure state.

Choose the correct statements from the following:

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 3

The correct answer is Option (3) → (B), (C) and (D) only

(A) two different solutions of sucrose of same molality prepared in different solvents will have the same depression in freezing point. — Incorrect. In \( \Delta T_f = K_f\,m \), the cryoscopic constant \( K_f \) is a property of the solvent (1.86 K kg mol⁻¹ for water, 5.12 K kg mol⁻¹ for benzene). The same molality in different solvents therefore gives different depressions.
(B) the osmotic pressure of a solution is given by the equation Π = CRT (where C is the molarity of the solution) — Correct. This is the van’t Hoff equation for a dilute solution: \( \Pi = \frac{n_2}{V}RT = CRT \).
(C) decreasing order of osmotic pressure for 0.01 M aqueous solution of \(BaCl_2, KCl, CH_3COOH\) and sucrose is \(BaCl_2> KCl > CH_3COOH\) > sucrose — Correct. At the same C and T, \( \Pi = iCRT \) is proportional to i:
    \( \displaystyle \ce{BaCl2 -> Ba^2+ + 2Cl-}:\ 1 + 2 = 3 \text{ ions},\ i = 3 \)
    \( \displaystyle \ce{KCl -> K+ + Cl-}:\ 2 \text{ ions},\ i = 2 \)
    \( \displaystyle \ce{CH3COOH <=> CH3COO- + H+}:\ \text{weak acid, only slightly ionised},\ i = 1 + \alpha \text{ (just above 1)} \)
    \( \displaystyle \text{Sucrose}:\ \text{does not dissociate},\ i = 1 \)
So \( \ce{BaCl2} \gt \ce{KCl} \gt \ce{CH3COOH} \gt \) sucrose.
(D) according to Raoult’s law the vapor pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution — Correct. Raoult’s law states \( p_1 = p_1^{0}\,x_1 \): the partial vapour pressure of each volatile component is proportional to its mole fraction, the constant being its vapour pressure in the pure state.