Question

Vapour pressure of a solution containing $6.0\text{g}$ of a nonvolatile solute in $180\text{g}$ water at $25°C$ is $20\text{torr}$. If 1 mole of water is further added, the vapour pressure at $25°C$ increases by $0.02\text{torr}$. Identify the correct information.

• A. The molar mass of solute is $54{\text{g mol}}^{-1}$
• B. The vapour pressure of pure water at ${25}^{o}C$ is $\frac{182}{9}\text{torr}$
• C. The molality of solution taken initially was $\frac{1}{1.62}m$
• D. The molality of final solution is $\frac{1}{1.782}m$

Answer 1

Answer: (A), (B), (C), (D)

The molality of final solution is $\frac{1}{1.782}m$

(a) Given solvent mass $\left(H_{2}O\right)\left(w_B\right)=180\text{g}$
Mass of non-volatile solute $=6g$
Vapour pressure $\left(P_s\right)=20\text{torr}$
According to Raoult’s law:
$\frac{P_A^o-P_s}{P_s}=\frac{n_B}{n_A}=\frac{w_B/M_B}{w_A/M_A}...\left(i\right)$
$\frac{P_A^o-20}{20}=\frac{6/M_B}{180/18}...\left(ii\right)$
When $1$ mole water was further added, the vapour pressure increased by $0.02\text{torr}$ then
$\frac{P_A^o-20.02}{20.02}=\frac{6/M_B}{10+1}...\left(iii\right)$
On equating equations (i) and (iii) we get
$\frac{P_A^o-20\times 20.02}{P_A^o-20.2\times 20}=\frac{11}{10}$
$10\times (P_A^o-20)\times 20.02=11\times (P_A^o-20.2)\times 20$
$20.2=0.99P_A^o$
(b) $P_A^o=\frac{2002}{99}=\frac{182}{9}\text{torr}$
(a) Molar mass of solute = $\left(M_B\right)$
From equation (i)
$\frac{\frac{182}{9}-120}{20}=\frac{6}{10}\times \frac{1}{M_B}$
$\frac{182-180}{9\times 20}=\frac{3}{5}\times \frac{1}{M_B}$
$\frac{2}{9\times 20}=\frac{3}{5}\times \frac{1}{M_B}$
$M_B=\frac{20\times 9\times 3}{2\times 5}=54\text{g/mol}$
(c) Initially molality $\left(m\right)=\frac{\text{mass of solute}}{\text{mass of solvent (kg)}}$
$=\frac{6/54}{180}\times 1000$
$=\frac{100}{9\times 18}=\frac{1}{1.62}m$
(d) Final molality $m=\frac{6/54}{198}\times 1000=\frac{1}{1.782}m$