Question

Lowering of vapour pressure due to a solute in 1 molal aqueous solution at ${100}^{\circ }C$ is

• A. 13.44 mm Hg
• B. 14.12 mm Hg
• C. 13.2 mm Hg
• D. 35.2 mm Hg

Answer 1

The correct option is A 13.44 mm Hg

We know that,
$M{w}_B=\frac{xB\times 1000}{(l-X_B)M{w}_A}$
$\left[\begin{array}{c}{X}_B=molefractionofsolute\\ M{W}_A=molemassofsolvent\end{array}\right]$
$1=\frac{x_B\times 1000}{(l-x_B)\times 18}$
$x_B=0.0176$
$x_A=1-0.0176=0.9824$
Now,
$P=P^{\circ }{x}_A=760\times 0.9824=746.62$
$\Delta P=P^{\circ }-P=760-746.62=13.41mmHg\left(approx\right)$