Question

Given that the freezing point of benzene is $ 5.48^\circ C $ and its $ K_f $ value is $ 5.12^\circ C/m $, what would be the freezing point of a solution of 20 g of propane in 400 g of benzene?

• A. \( -0.34^\circ C \)
• B. \( -0.17^\circ C \)
• C. \( -5.8^\circ C \)
• D. \( -0.2^\circ C \)

Hint:

To calculate freezing point depression, remember that the formula involves the molality of the solution, which is based on the moles of solute and the mass of solvent.

Answer 1

The Correct Option is A

To calculate the freezing point depression (\(\Delta T_f\)) of the solution, we use the formula: \[ \Delta T_f = K_f \times m \] where: - \( K_f \) is the freezing point depression constant (given as \( 5.12^\circ C/m \)), - \( m \) is the molality of the solution. First, we calculate the molality \( m \) of the solution. The molality is given by the formula: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] 1. Calculate the moles of propane (C\(_3\)H\(_8\)): - Molar mass of propane (C\(_3\)H\(_8\)) = \( 3 \times 12 + 8 \times 1 = 44 \, \text{g/mol} \) - Moles of propane = \( \frac{20 \, \text{g}}{44 \, \text{g/mol}} = 0.4545 \, \text{mol} \) 2. Calculate the mass of benzene in kg: - Mass of benzene = 400 g = \( 0.4 \, \text{kg} \) 3. Calculate the molality \( m \): \[ m = \frac{0.4545 \, \text{mol}}{0.4 \, \text{kg}} = 1.13625 \, \text{mol/kg} \] 4. Calculate the freezing point depression \( \Delta T_f \): \[ \Delta T_f = 5.12^\circ C/m \times 1.13625 \, \text{mol/kg} = 5.81^\circ C \] 5. Calculate the new freezing point: - Freezing point of pure benzene = \( 5.48^\circ C \) - New freezing point = \( 5.48^\circ C - 5.81^\circ C = -0.34^\circ C \) Thus, the freezing point of the solution is \( -0.34^\circ C \).