The correct option is Carbon tetrachloride.
Explanation
The depression in freezing point ($\Delta T_f$) is a colligative property of solutions, which depends on the number of solute particles rather than their identity. Mathematically, it is expressed as:
$\Delta T_f = K_f \times m$
Where:
- $K_f$ is the molal depression constant (or cryoscopic constant) specific to the solvent.
- $m$ is the molality of the solution.
Analysis:
The question states that the molality ($m$) of the solute is the same for all solvents. Therefore, the depression in freezing point is directly proportional to the cryoscopic constant ($K_f$) of the solvent ($\Delta T_f \propto K_f$). To determine which solvent shows the largest depression, one must compare their $K_f$ values:
- Water: $K_f \approx 1.86 \, \text{K kg mol}^{-1}$
- Ethanol: $K_f \approx 1.99 \, \text{K kg mol}^{-1}$
- Diethyl ether: $K_f \approx 1.79 \, \text{K kg mol}^{-1}$
- Carbon tetrachloride ($CCl_4$): $K_f \approx 29.8 \, \text{K kg mol}^{-1}$
Carbon tetrachloride has a significantly higher $K_f$ value compared to water, ethanol, and diethyl ether. Consequently, for the same molal concentration of a solute, the solution using carbon tetrachloride as the solvent will exhibit the largest decrease in freezing point.
Key Takeaway:
For equimolal solutions, the magnitude of the depression in freezing point is determined solely by the cryoscopic constant ($K_f$) of the solvent. Solvents with higher $K_f$ values, such as carbon tetrachloride or camphor, show larger freezing point depressions.