Question

A solution is prepared by dissolving $10 \,g$ of a non-volatile solute (molar mass, $'M^{\prime} g mol ^{-1}$ ) in $360\, g$ of water. What is the molar mass in $g\, mol ^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$ ?

• A. 199
• B. 99.5
• C. 299
• D. 149.5

Answer 1

Answer: (B)

Given,

Mass of solute $\left(w_{B}\right)=10 \,g$

Molar mass of solute $\left(M_{B}\right)=M_{B}$

Mass of solvent $\left(w_{A}\right)=360 \,g$

Relative lowering in vapour pressure of solution

$=5 \times 10^{-3}$

Molar mass of water $\left(M_{A}\right)=18 \,g \,mol ^{-1}$

$\because \frac{\Delta p}{p^{\circ}}=$ Relative lower in vapour-pressure of solution.

$\frac{\Delta p}{p^{\circ}}=\chi_{B} $

$\Rightarrow \frac{n_{B}}{n_{A}+n_{B}}=5 \times 10^{-3}$

where, $n_{A}$ and $n_{B}$ are number of moles of solvent

(A) and solute $(B)$ respectively.

$n_{A}=\frac{360}{18}=20$

$n_{B}=\frac{w_{B}}{M_{B}}=\frac{10}{M_{B}}$

$ \because 5 \times 10^{-3}=\chi_{B}=\frac{\frac{10}{M_{B}}}{20+\frac{10}{M_{B}}}$

$5 \times 10^{-3}=\frac{\frac{10}{M_{B}}}{\frac{20 M_{B}+10}{M_{B}}}$

$5 \times 10^{-3}=\frac{10}{20\, M_{B}+10}$

$\left(20\, M_{B}+10\right) 5 \times 10^{-3} =10 $ or,

$20\, M_{B}+10 =\frac{10}{5} \times 10^{3} $

$20 M_{B}+10 =2000 $

$\Rightarrow 20 \,M_{B}=1990 $

$ M_{B} =\frac{1990}{20}=99.5 \,g \,mol ^{-1}$