Question

An aqueous solution containing 10 gm of starch per litre has osmotic pressure of 3.8 mm of Hg at 27°C. Then :

[Given : Kf = 1.86°C / molal, vapour pressure of pure water at 27°C = 20 torr]

• A. Molar mass of starch is 48 kg
• B. Freezing point of solution is slightly more than freezing point of pure solvent
• C. Vapour pressure of solution is slightly less than vapour pressure of pure solvent
• D. On addition of starch boiling point will increase.

Answer 1

Answer: (A), (C), (D)

A, C, and D

Answer 2

1. Molar mass determination using osmotic pressure:

The osmotic pressure formula is

$$\pi = MRT,$$

where $\pi = 3.8\; \text{mmHg}$ ( = 3.8 torr), $R = 0.08206\; \text{L·atm/mol·K}$, and $T = 27^\circ\text{C} + 273 = 300\; \text{K}$.

First, convert $\pi$ to atm:

$$\pi = \frac{3.8}{760}\; \text{atm} \approx 0.0050\; \text{atm}.$$

Now, the molarity $M$ is

$$M = \frac{\pi}{RT} = \frac{0.0050}{0.08206 \times 300} \approx 0.0002033\; \text{mol/L}.$$

Since the solution contains $10\, \text{g}$ of starch per litre, the molar mass $M_m$ of starch is

$$M_m = \frac{10\, \text{g}}{0.0002033\, \text{mol}} \approx 49200\, \text{g/mol} \approx 49.2\, \text{kg/mol},$$

which is close to the given option (A) stating 48 kg.

2. Freezing point:

Freezing point depression is given by

$$\Delta T_f = K_f \, m,$$

where $K_f = 1.86^\circ\text{C/m}$. The presence of starch (a non-volatile solute) will lower the freezing point of water. Thus, the freezing point of the solution will be lower, not higher than that of pure solvent. Option (B) is false.

3. Vapour pressure:

According to Raoult's law, the vapour pressure of the solvent over a solution is less than that of the pure solvent. Thus, option (C) is correct.

4. Boiling point:

The addition of a non-volatile solute (starch) elevates the boiling point of the solution (boiling point elevation). Therefore, option (D) is correct.