Question

$1.355g$ of a substance dissolved in $55g$ of $C{H}_{3}COOH$ produced a depression in the freezing point of ${0.618}^{\circ }C$. Calculate the molar mass of the substance. $(K_f={3.85}^{\circ }Ckgmo{l}^{-1})$

• A. $160g/mol$
• B. $153.5g/mol$
• C. $133.7g/mol$
• D. $141.5g/mol$

Answer 1

The correct option is B $153.5g/mol$

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.
From the above equation,
We know that,
$\Delta T=K_f\times \frac{w\times 1000}{M\times W}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute
Given:
$\text{Weight of solute,}w=1.355g$
$\text{Weight of solvent,}W=55g$
$\text{Depression in freezing point,}△T_f={0.618}^{\circ }C$
$\text{Molal depression constant,}{K}_f={3.85}^{\circ }Ckgmo{l}^{-1}$
Substituting the given values in above equation gives,
$0.618=3.85\times \frac{1.355\times 1000}{M\times 55}M=153.47\approx 153.5g/mol$