Question

A sample of benzene, contaminated with a non-volatile and non-ionic solute, boils at 0.31°C higher than that of pure benzene. The molality of the solute in the contaminated solution is ________ (round off to two decimal places).

Hint:

The boiling point elevation is directly proportional to the molality of the solute. Use the ebullioscopic constant to calculate the molality in such problems.

Answer 1

Correct Answer: 0.11

The boiling point elevation \( \Delta T_b \) is given by the formula: \[ \Delta T_b = K_b \cdot m \] Where:
- \( \Delta T_b = 0.31^\circ \text{C} \) is the boiling point elevation,
- \( K_b \) is the ebullioscopic constant of the solvent (benzene in this case),
- \( m \) is the molality of the solute.
The value of \( K_b \) for benzene can be calculated using the following formula: \[ K_b = \frac{R \cdot T_b^2}{\Delta H_{\text{vap}}} \] Where:
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \) is the gas constant,
- \( T_b = 80.1^\circ \text{C} = 353.25 \, \text{K} \) is the boiling point of benzene,
- \( \Delta H_{\text{vap}} = 30.76 \, \text{kJ/mol} = 30760 \, \text{J/mol} \) is the enthalpy of vaporization.
Substituting the values into the formula for \( K_b \): \[ K_b = \frac{8.314 \cdot (353.25)^2}{30760} \approx 2.53 \, \text{K kg/mol}. \] Now, using the boiling point elevation formula: \[ 0.31 = 2.53 \cdot m \] Solving for \( m \): \[ m = \frac{0.31}{2.53} \approx 0.12 \, \text{mol/kg}. \] Thus, the molality of the solute in the contaminated solution is approximately \( \boxed{0.12} \, \text{mol/kg} \).