The depression of freezing point is given by the equation: \[ \Delta T_f = K_f \cdot m \] Where:
\( \Delta T_f \) is the depression in freezing point,
\( K_f \) is the freezing point depression constant, and
\( m \) is the molality of the solution.
Step 1: Solve for molality To find the molality \( m \), we can rearrange the formula: \[ m = \frac{\Delta T_f}{K_f} \] This equation shows that molality is directly proportional to the depression in freezing point and inversely proportional to the freezing point depression constant.
Thus, the correct answer is \( {m = \frac{\Delta T_f}{K_f}} \).
Given below are two statements:
Statement (I): Molal depression constant $ k_f $ is given by $ \frac{M_1 R T_f}{\Delta S_{\text{fus}}} $, where symbols have their usual meaning.
Statement (II): $ k_f $ for benzene is less than the $ k_f $ for water.
In light of the above statements, choose the most appropriate answer from the options given below: