Question

5 g of a solute X (M. wt = 200 g/mol) is dissolved in 250 g benzene. If \( \Delta T_f = 0.5 \, \text{K} \) and the relative lowering of vapour pressure is \( P \times 10^4 \), find \( P \). Given: \( K_f = 5.5 \, \text{K} \, \text{kg} \, \text{mol}^{-1} \), solute dimerises in benzene.

• A. 253.6
• B. 0.1636
• C. 70
• D. 23.36

Hint:

When the solute undergoes dimerisation, the van't Hoff factor \( i \) is adjusted accordingly to reflect the number of particles formed in solution. For dimerisation, \( i \) is typically less than 1.

Answer 1

The Correct Option is C

Step 1: Use the formula for freezing point depression.
The freezing point depression is related to the molality of the solution by the formula: \[ \Delta T_f = i \times K_f \times m \] where: - \( \Delta T_f \) is the freezing point depression, - \( i \) is the van't Hoff factor (which accounts for the dissociation of the solute), - \( K_f \) is the cryoscopic constant of the solvent, - \( m \) is the molality of the solution.
Step 2: Determine the molality.
First, calculate the molality \( m \): \[ m = \frac{\text{mol of solute}}{\text{mass of solvent (kg)}} \] The mass of solute is 5 g, and the molar mass of the solute is 200 g/mol. Therefore, the number of moles of solute is: \[ \text{mol of solute} = \frac{5}{200} = 0.025 \, \text{mol} \] The mass of the solvent (benzene) is 250 g, which is 0.250 kg. Thus, the molality \( m \) is: \[ m = \frac{0.025}{0.250} = 0.1 \, \text{mol/kg} \]
Step 3: Use the given \( \Delta T_f \) and \( K_f \).
We are given \( \Delta T_f = 0.5 \, \text{K} \), and \( K_f = 5.5 \, \text{K} \, \text{kg/mol} \). Using the formula for freezing point depression: \[ 0.5 = i \times 5.5 \times 0.1 \] Solving for \( i \): \[ i = \frac{0.5}{5.5 \times 0.1} = \frac{0.5}{0.55} = 0.909 \] Since \( i \) represents the number of particles formed in solution, this suggests that the solute undergoes dimerisation (as \( i \) is less than 1, indicating that the solute molecules combine to form dimers).
Step 4: Calculate the relative lowering of vapour pressure.
The relative lowering of vapour pressure is given by: \[ \frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{i \times \text{mol of solute}}{\text{mol of solvent}} \] The molar mass of benzene is 78 g/mol. The number of moles of benzene is: \[ \text{mol of benzene} = \frac{250}{78} = 3.205 \, \text{mol} \] Thus, the relative lowering of vapour pressure is: \[ \frac{\Delta P}{P_0} = \frac{0.909 \times 0.025}{3.205} = 0.0071 \] Finally, the lowering of pressure \( \Delta P = P_0 \times 0.0071 \), and since \( \Delta P = P \times 10^{-4} \), we find: \[ P = 0.0071 \times 10^4 = 70 \] Thus, the value of \( P \) is \( 70 \times 10^{-4} \).