Question

$1.5g$ of an unknown substance was dissolved in $7.5g$ of camphor and it was found that the melting point of camphor was depressed by $5.3K$. If $K_f$ is $39.75Kkgmo{l}^{-1}$, find the molecular mass of the solute.

• A. $475g/mol$
• B. $600g/mol$
• C. $1500g/mol$
• D. $300g/mol$

Answer 1

The correct option is C $1500g/mol$

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

$\text{Given,}{K}_f=39.75Kkgmo{l}^{-1}w=1.5g,W=7.5g,△T_f=5.3K$

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


$M=\frac{39.75\times 1.5\times 1000}{7.5\times 5.3}M=1500g/mol$