Question

By dissolving $13.6g$ of a substance in $20g$ of water, the freezing point decreased by ${3.7}^{\circ }C$. Calculate the molecular mass of the substance. (Molal depression constant for water = $1.863Kkgmo{l}^{-1}$)

• A. $242.4g/mol$
• B. $432.4g/mol$
• C. $342.4g/mol$
• D. $243.4g/mol$

Answer 1

The correct option is C $342.4g/mol$

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.
$\text{Molality}=\frac{△T_f}{K_f}$
$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
$\therefore$
$\frac{△T_f}{K_f}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute
$M=\frac{1000K_f\times w}{W\times △T_f}\text{Given,}{K}_f=1.863Kkgmo{l}^{-1}w=13.6g,W=20g,△T_f={3.7}^{\circ }CM=\frac{1000\times 1.863\times 13.6}{20\times 3.7}\Rightarrow 342.39$
Molar mass of the substance is $342.4g/mol$