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

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

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Solution:

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=X_{2}$ (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 $=0.9868 \times 18=17.76\, g$

(molecular weight of $\left. H _{2} O =18\right)$

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