Question

A solution containing 30 g of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mmHg at $25^{\circ}C$. Further 18 g of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at $25^{\circ}C$ Calculate the molecular weight of the solute

• A. $74.2$
• B. $75.6$
• C. $67.83$
• D. $78.7$

Answer 1

Answer: (C)

We have,
$\frac{P^{o}-21.85}{21.85}=\frac{30\times18}{90\times m}$, for I case $\ldots\ldots\left(i\right)$
wt. of solvent $= 90 + 18= 108 \,gm$
$\frac{P^{o}-22.15}{22.15}=\frac{30\times18}{108\times m}$, for II case $\ldots\ldots\left(ii\right)$
By e q .$\left(1\right)P_{m}^{o}-21.85\,m=21.85\times3=131.1$
By eq. $\left(2\right) P_{m}^{o}-22.15\,m=22.15\times5=110.75$
$0.30\,m=20.35$
$m=\frac{20.35}{0.30}=67.83$