Question

On dissolving $0.25g$ of a non volatile substance in $30mL$ of benzene (Density of benzene $0.8g/mL$) its freezing point decreases by ${0.40}^{\circ }C$. Calculate the molar mass of non - volatile substance $(K_f=5.12Kkgmo{l}^{-1})$

• A. $133.33g/mol$
• B. $163.5g/mol$
• C. $333.54g/mol$
• D. $201.6g/mol$

Answer 1

The correct option is A $133.33g/mol$

$\text{Mass of benzene,}W=\text{Volume}\times \text{Density}$
$=30\times 0.8=24g$
Given,
$\text{Cryoscopic constant,}{K}_f=5.12Kkgmo{l}^{-1}\text{Weight of solute,}w=0.25g\Delta T={0.40}^{\circ }C.$

Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
m is molality of the solution.
$\text{Molality}=\frac{△T_f}{K_f}$
$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
$M$ is molar mass of the solute
$\therefore$
$\frac{△T_f}{K_f}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
Thus,
$M=\frac{1000K_f\times w}{W\times \Delta T}=\frac{1000\times 5.12\times 0.25}{24\times 0.40}=133.33g/mol$