Question

The freezing point of a solution containing $2.40$ g of a compound in $60.0$ g of benzene is $0.10$${}^o$C lower than that of pure benzene. What is the molecular weight of the compound?

[$K$ is $5.12$${}^o$C/m for benzene]

• A. $2000g/mol$
• B. $2050g/mol$
• C. $2100g/mol$
• D. $2150g/mol$

Answer 1

The correct option is B $2050g/mol$

The expression for the depression in the freezing point and the molar mass of the solute is as shown below.

$\Delta T_f=\frac{K_f\times W_2}{M_2\times W_1}$

Substitute values in the above expression.

$0.1=\frac{5.12\times 2.40}{M_2\times 60}$

Thus the molecular weight of the compound is 2.050 kg/mol or 2050 g/mol.

Hence, the correct option is $\text{B}$