Question

The depression in freezing point of a solution of molality \(0.01\ mol\ kg^{-1}\) is highest with respect to which of the following solvents? (Values of \(K_f\) are given in brackets)

• A. Water \((1.86)\)
• B. Benzene \((5.12)\)
• C. Carbon tetrachloride \((31.8)\)
• D. Cyclohexane \((20.0)\)

Hint:

For solutions having the same molality, \[ \Delta T_f = K_fm \] Hence, the solvent with the largest \(K_f\) value exhibits the greatest depression in freezing point.

Answer 1

The Correct Option is C

Step 1: Recall the formula for depression in freezing point.
The depression in freezing point is given by \[ \Delta T_f = iK_fm \] where \[ i=\text{van't Hoff factor} \] \[ K_f=\text{cryoscopic constant} \] \[ m=\text{molality} \]

Step 2: Analyze the given data.
The molality is the same for all solutions: \[ m=0.01\ mol\ kg^{-1} \] Since the same solute concentration is considered, \(i\) and \(m\) remain constant.
Therefore, \[ \Delta T_f \propto K_f \]

Step 3: Compare the \(K_f\) values.
Given, \[ K_f(\text{Water})=1.86 \] \[ K_f(\text{Benzene})=5.12 \] \[ K_f(\text{Cyclohexane})=20.0 \] \[ K_f(\text{Carbon tetrachloride})=31.8 \] Among these values, \[ 31.8 \] is the highest.
Therefore, carbon tetrachloride will show the maximum depression in freezing point.

Step 4: Verify numerically.
For carbon tetrachloride, \[ \Delta T_f = K_fm \] \[ =31.8\times 0.01 \] \[ =0.318^\circ C \] This is greater than the corresponding values for all other solvents.

Step 5: Final conclusion.
Hence, the highest depression in freezing point is obtained with \[ \boxed{\text{Carbon tetrachloride}} \] Therefore, the correct option is (3).