Question

On dissolution of 2.0 g of a non-electrolyte in 50.0 g benzene, its freezing point decreases by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol$^{-1}$. Calculate the molar mass of the solute.

Hint:

Freezing point depression can be used to calculate the molar mass of a solute when the solvent’s freezing point depression constant is known.

Answer 1

Step 1: Formula for freezing point depression. 
The freezing point depression \( \Delta T_f \) is related to the molality \( m \) by the formula: \[ \Delta T_f = K_f \times m \] where \( K_f \) is the freezing point depression constant and \( m \) is the molality. 

Step 2: Calculate molality. 
We are given that \( \Delta T_f = 0.40 \, \text{K} \) and \( K_f = 5.12 \, \text{K kg mol}^{-1} \), so: \[ m = \frac{\Delta T_f}{K_f} = \frac{0.40}{5.12} = 0.0781 \, \text{mol/kg} \] 

Step 3: Calculate molar mass. 
Molality is given by: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \] The mass of the solvent (benzene) is 50.0 g = 0.0500 kg. Thus: \[ \text{Moles of solute} = m \times \text{mass of solvent} = 0.0781 \times 0.0500 = 0.003905 \, \text{mol} \] 

Step 4: Calculate molar mass. 
The molar mass is given by: \[ \text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.0 \, \text{g}}{0.003905 \, \text{mol}} = 160.25 \, \text{g/mol} \] 

Step 5: Conclusion. 
Hence, the correct answer is (B) 160 g mol$^{-1}$.