Question

A dilute aqueous solution of glucose shows a vapour pressure of $750mm$ of $Hg$ at $373K$. The molality of the solution is

• A. 13.32
• B. 0.013
• C. 1.35
• D. 0.74

Answer 1

Answer: (D)

$\frac{p^{\circ }-p_s}{p^{\circ }}=X_2$ (Raoult's law)

$\frac{760-750}{760}=0.0132$

Mole fraction of solvent $\left(X_1\right)=1-0.0132=0.9868$

It means $0.0132$ mole of glucose is present per $0.9868$ mole of water.

Weight of solvent $=0.9868\times 18=17.76g$

(molecular weight of $H_{2}O=18)$

Molality $=\frac{0.0132\times 1000}{17.76}=0.74$