Question

A solution is prepared by dissolving $10g$ of a non-volatile solute (molar mass, ${}^{\prime }{M}^{\prime }gmo{l}^{-1}$ ) in $360g$ of water. What is the molar mass in $gmo{l}^{-1}$ 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 $\left(w_B\right)=10g$

Molar mass of solute $\left(M_B\right)=M_B$

Mass of solvent $\left(w_A\right)=360g$

Relative lowering in vapour pressure of solution

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

Molar mass of water $\left(M_A\right)=18gmo{l}^{-1}$

$\because \frac{\Delta p}{p^{\circ }}=$ 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.

$n_A=\frac{360}{18}=20$

$n_B=\frac{w_B}{M_B}=\frac{10}{M_B}$

$\because 5\times 10^{-3}={\chi }_B=\frac{\frac{10}{M_B}}{20+\frac{10}{M_B}}$

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

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

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

$20M_B+10=\frac{10}{5}\times 10^3$

$20M_B+10=2000$

$\Rightarrow 20M_B=1990$

$M_B=\frac{1990}{20}=99.5gmo{l}^{-1}$