Question

A solution containing $0.5216g$ of non-volatile solute 'A' (molar wt. = $128.16gmo{l}^{-1}$) in $50mL$ of $CC{l}_4$ shows boiling-point elevation of ${0.402}^{o}C$ while a solution of $0.6216g$ of an unknown solute in the same weight of solvent gave a boiling-point elevation of ${0.647}^{o}C.$ Find the molecular mass of the unknown solute.

• A. $130g/mol$
• B. $99.5g/mol$
• C. $85.5g/mol$
• D. $94.84g/mol$

Answer 1

The correct option is D $94.84g/mol$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.

$\text{Mass of the solute 'A' in first solution}=0.5216g$
$\text{No. of moles of solute 'A' in first solution}=\frac{0.5216}{128.16}$
Suppose the weight of $50mL$ of solvent $CC{l}_4$ is $Wg$.
For the first solution,
Molality $=\frac{0.5216/128.16}{W}\times 1000$
$=\frac{521.6}{128.16W}$
$K_b=\frac{△T_b}{m}$
$=\frac{0.402}{521.6/128.16W}$
As for the second solution,
Molality $=\frac{0.6216/M}{W}\times 1000$
$=\frac{621.6}{WM}$
$\therefore K_b^{\prime }=\frac{0.647}{621.6/WM}$
($M$ is the molar weight of the unknown solute in second solution.)

Since, $K_b$ is the property of the solvent and is independent of the solute.
For the same solvent, $K_b=K_b^{\prime }$
Hence, $\frac{0.402}{521.6/128.16W}=\frac{0.647}{621.6/WM}$
$M=94.84g/mol$