Question

Given below are two statements:

Statement (I): Molal depression constant $K_f$ is given by $\frac{M_1RT_f}{\Delta S_{fus}}$, where symbols have their usual meaning.

Statement (II) : $K_f$ for benzene is less than the $K_f$ for water.

In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Both Statement (I) and Statement (II) are true
• B. Both Statement (I) and Statement (II) are false
• C. Statement (I) is true but Statement (II) is false
• D. Statement (I) is false but Statement (II) is true

Answer 1

Answer: (C)

Statement (I) is true but Statement (II) is false

Answer 2

Statement I Verification:

The freezing point depression is given by:

$$\Delta T_f = K_f \cdot m$$

From thermodynamics, we can derive:

$$\Delta T_f = \frac{R \cdot T_f^2}{\Delta H_{fus}} \cdot m$$

Since $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$, we have:

$$K_f = \frac{R \cdot T_f^2}{\Delta H_{fus}} = \frac{R \cdot T_f^2}{T_f \cdot \Delta S_{fus}} = \frac{R \cdot T_f}{\Delta S_{fus}}$$

When expressed in terms of the solvent’s molar mass $M_1$ (in kg/mol) for unit consistency of $K_f$ in °C·kg/mol:

$$K_f = \frac{M_1 \cdot R \cdot T_f}{\Delta S_{fus}}$$

Thus, Statement (I) is correct.

Statement II Verification:

For water:

• $T_f = 273$ K
• $M_1 \approx 0.018$ kg/mol
• $\Delta H_{fus} \approx 6010$ J/mol $\implies$ $\Delta S_{fus} \approx \frac{6010}{273} \approx 22$ J/(mol·K)

Calculating $K_f$ for water:

$$K_f = \frac{0.018 \times 8.314 \times 273}{22} \approx 1.85 \frac{°C \cdot kg}{mol}$$

For benzene:

• $T_f \approx 278.65$ K
• $M_1 \approx 0.078$ kg/mol
• $\Delta H_{fus} \approx 9900$ J/mol $\implies$ $\Delta S_{fus} \approx \frac{9900}{278.65} \approx 35.54$ J/(mol·K)

Then,

$$K_f = \frac{0.078 \times 8.314 \times 278.65}{35.54} \approx 5.08 \frac{°C \cdot kg}{mol}$$

Since $K_f$ for benzene ($\approx 5.08$) is greater than that for water ($\approx 1.85$), Statement (II) is false.