A mixture of an organic liquid $A$ and water distilled under one atmospheric pressure at $99. 2^{o} C$ . How many grams of steam will be condensed to obtained $1.0 g$ of liquid $A$ in the distillate? (Vapour pressure of water at ${99.2}^{o} C$ is $739 m m H g$ . Molecular weight of $A = 123$ )

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Solution

Total pressure of the solution(A+Water) $= 1$ atm $= 760$ mm Hg
Vapour pressure of water $= 739$ mm Hg
Vapour pressure of organic liquid A is $= 760 - 739 = 21$ mm Hg
Moleucular mass of A $= 123$
$\frac{W_{A}}{W_{w a t e r}} = \frac{P_{A} \times M M_{A}}{P_{B} \times M M_{w} a t e r}$
Where $W_{A}$ & $W_{w a t e r}$ represents the mass of organic liquid A and mass of water respectively.
$M M_{A}$ & $M M_{w} a t e r$ represents the molar mass of A and molar mass of water respectively
$\frac{1}{W_{w a t e r}} = \frac{21 \times 123}{739 \times 18} = 5.19$ g

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