Question

Addition of 0.643g of a compound to 50mL of benzene (density = 0.879gmL$^{-1}$), lowers the freezing point from 50.51°C to 50.03°C. If $K_f$ for benzene is 5.12Kkgmol$^{-1}$, the molecular mass (in gmol$^{-1}$) of the compound is

• A. 156.05
• B. 312.00
• C. 78.00
• D. 468.00

Answer 1

Answer: (A)

156.05 gmol$^{-1}$

Answer 2

To find the molecular mass of the compound, we use the colligative property of freezing point depression. The formula relating the freezing point depression ($\Delta T_f$) to the molality ($m$) and the cryoscopic constant ($K_f$) is:

$\Delta T_f = K_f \cdot m$

Where:

• $\Delta T_f$ is the freezing point depression.
• $K_f$ is the cryoscopic constant of the solvent.
• $m$ is the molality of the solution.

Molality is defined as moles of solute per kilogram of solvent:

$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$

The number of moles of solute can be expressed as:

$\text{moles of solute} = \frac{W_B}{M_B}$

Where:

• $W_B$ is the mass of the solute.
• $M_B$ is the molecular mass of the solute.

The mass of the solvent ($W_A$) in kilograms is:

$\frac{W_A \text{ (in g)}}{1000}$

So, the molality can be rewritten as:

$m = \frac{W_B/M_B}{W_A/1000} = \frac{W_B \times 1000}{M_B \times W_A}$

Substituting this into the freezing point depression formula:

$\Delta T_f = K_f \times \frac{W_B \times 1000}{M_B \times W_A}$

Given:

• Mass of solute, $W_B = 0.643$ g
• Volume of benzene (solvent), $V_A = 50$ mL
• Density of benzene, $\rho_A = 0.879$ g/mL
• Freezing point of pure benzene, $T_f^0 = 50.51^\circ$C
• Freezing point of solution, $T_f = 50.03^\circ$C
• Cryoscopic constant for benzene, $K_f = 5.12$ K kg mol$^{-1}$

First, calculate the mass of benzene ($W_A$) using its volume and density:

$W_A = \rho_A \times V_A = 0.879 \text{ g/mL} \times 50 \text{ mL} = 43.95$ g

Next, calculate the freezing point depression ($\Delta T_f$):

$\Delta T_f = T_f^0 - T_f = 50.51^\circ\text{C} - 50.03^\circ\text{C} = 0.48^\circ\text{C} = 0.48$ K

Now, rearrange the formula $\Delta T_f = \frac{K_f \times W_B \times 1000}{M_B \times W_A}$ to solve for $M_B$:

$M_B = \frac{K_f \times W_B \times 1000}{\Delta T_f \times W_A}$

Substitute the known values into the formula:

$M_B = \frac{5.12 \text{ K kg mol}^{-1} \times 0.643 \text{ g} \times 1000 \text{ g/kg}}{0.48 \text{ K} \times 43.95 \text{ g}}$

$M_B = \frac{5.12 \times 0.643 \times 1000}{0.48 \times 43.95} = \frac{3292.16}{21.096} \approx 156.05$ g/mol