Question

10g of a nonvolatile solute, when dissolved in 100g of benzene, raises its boiling point by \(1^\circ C\). The molecular mass of the solute in g/mol is (K_b for benzene = 2.53 K·kg/mol):

• A. 25.3
• B. 253
• C. 250
• D. 25.0

Hint:

To find the molecular mass using boiling point elevation, use the formula \( \Delta T_b = K_b \times m \), where molality \( m \) is calculated as \( \frac{\text{moles of solute}}{\text{kg of solvent}} \).

Answer 1

The Correct Option is A

Step 1: Boiling point elevation formula.
The formula for boiling point elevation is: \[ \Delta T_b = K_b \times m \] where \( \Delta T_b \) is the change in boiling point, \( K_b \) is the ebullioscopic constant (given as \( 2.53 \, \text{K·kg/mol} \) for benzene), and \( m \) is the molality of the solution.

Step 2: Calculate the molality.
We are given that \( \Delta T_b = 1^\circ C \), and we know \( K_b = 2.53 \, \text{K·kg/mol} \). Thus: \[ 1 = 2.53 \times m \] Solving for \( m \): \[ m = \frac{1}{2.53} = 0.395 \, \text{mol/kg} \]

Step 3: Calculate the molecular mass.
Molality is also defined as: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] Given that 10g of solute is dissolved in 100g (0.1 kg) of benzene, the number of moles of solute is: \[ \text{moles of solute} = m \times \text{kg of solvent} = 0.395 \times 0.1 = 0.0395 \, \text{mol} \] Now, the molecular mass \( M \) of the solute is: \[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{10}{0.0395} = 253.2 \, \text{g/mol} \] Thus, the molecular mass is approximately \( 25.3 \times 10^1 \, \text{g/mol} \).

Step 4: Conclusion.
The correct answer is option (1) \( 25.3 \, \text{g/mol} \).