Question

The freezing point depression of a solution containing \(0.6\ \text{g}\) of urea \((\text{molar mass}=60\ \text{g mol}^{-1})\) in \(100\ \text{g}\) of benzene is in K \((K_f=4.0\ \text{K kg mol}^{-1})\):

• A. \(0.30\)
• B. \(0.58\)
• C. \(0.40\)
• D. \(0.24\)

Hint:

For non-electrolytes like urea, \[ \Delta T_f=K_fm \] where molality is calculated using mass of solvent in kg, not total solution mass.

Answer 1

The Correct Option is C

Step 1: Write the formula for depression in freezing point.
Depression in freezing point is given by \[ \Delta T_f=K_fm \] where \[ K_f=\text{cryoscopic constant} \] and \[ m=\text{molality} \]

Step 2: Calculate moles of urea.
Given mass of urea: \[ w=0.6\ \text{g} \] Molar mass of urea: \[ M=60\ \text{g mol}^{-1} \] Therefore, \[ \text{Moles of urea}=\frac{0.6}{60} \] \[ =0.01\ \text{mol} \]

Step 3: Convert mass of solvent into kg.
Mass of benzene: \[ 100\ \text{g} \] Since, \[ 1000\ \text{g}=1\ \text{kg} \] Therefore, \[ 100\ \text{g}=0.1\ \text{kg} \]

Step 4: Calculate molality.
\[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] \[ m=\frac{0.01}{0.1} \] \[ m=0.1\ \text{mol kg}^{-1} \]

Step 5: Calculate depression in freezing point.
Given, \[ K_f=4.0\ \text{K kg mol}^{-1} \] Therefore, \[ \Delta T_f=K_fm \] \[ \Delta T_f=4.0\times0.1 \] \[ \Delta T_f=0.40\ K \]

Step 6: Final conclusion.
Therefore, the freezing point depression is \[ \boxed{0.40\ K} \] Hence, the correct option is \[ \boxed{(3)} \]