Question

Addition of $0.6g$ of a compound to $50mL$ of benzene $(\text{density}=0.88g/mL$) lowers the freezing point from ${5.51}^{o}Cto{5.01}^{o}C$. If $K_f$ for benzene is $5.1Kkgmo{l}^{-1}$, calculate the molecular weight of the compound.

• A. $90g/mol$
• B. $54g/mol$
• C. $139g/mol$
• D. $165g/mol$

Answer 1

The correct option is C $139g/mol$

Given:
$K_f=5.12Kkgmo{l}^{-1}$

$\text{Weight of solute,}w=0.6g$

$\Delta T_f=T_0-T_1=5.51-5.01=0.5$

$\text{Weight of benzene, W = Volume}\times \text{Density}$
$W=50mL\times 0.88$
$W=44g$

Freezing point depression of a solution is given by,
$△T_f=K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of the solution.
$\therefore$
$\frac{△T_f}{K_f}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$M=\frac{K_f\times w\times 1000}{\Delta T_f\times W}$

$M=\frac{5.1\times 0.6\times 1000}{0.5\times 44}$

$M=139.1g/mol$
$M\approx 139g/mol$