Given below are two statements :
Statement (I) : Molal depression constant $$K_f$$ is given by $$\frac{M_1 R T_f}{\Delta S_{fus}}$$, 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 :

We need to evaluate two statements about the molal depression constant $$K_f$$.

Statement I: Molal depression constant $$K_f$$ is given by $$\frac{M_1 R T_f}{\Delta S_{fus}}$$

The standard expression for the molal freezing point depression constant is:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times \Delta H_{fus}}$$$

where $$M_1$$ is the molar mass of the solvent, $$R$$ is the gas constant, $$T_f$$ is the freezing point of the solvent, and $$\Delta H_{fus}$$ is the enthalpy of fusion.

Since the entropy of fusion is defined as:

$$$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$$$

We can substitute $$\Delta H_{fus} = T_f \times \Delta S_{fus}$$:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times T_f \times \Delta S_{fus}} = \frac{M_1 R T_f}{1000 \times \Delta S_{fus}}$$$

When $$M_1$$ is expressed in kg/mol (i.e., $$M_1$$ already includes the factor of 1000), the expression simplifies to:

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

Therefore, Statement I is correct.

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

The known values of $$K_f$$ are:

$$K_f$$ for water $$= 1.86 \text{ K kg mol}^{-1}$$

$$K_f$$ for benzene $$= 5.12 \text{ K kg mol}^{-1}$$

Clearly, $$K_f$$ for benzene $$(5.12)$$ is greater than $$K_f$$ for water $$(1.86)$$.

Therefore, Statement II is incorrect.

Statement I is correct but Statement II is incorrect.

Hence, the correct answer is Option D.