A mixture of $N_{2}$ and water vapours is admitted to a flask which contains a solid drying agent. Immediately after admission, the pressure of the flask is 760 mm. After standing some hours, the pressure reached a steady value of 745 mm.
Calculate the composition of $N_{2}$ in mole % of original mixture (as nearest integer)

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Solution

Given, $P_{N_{2}}^{'} + P_{H_{2} O}^{'} = 760 m m$
After the use of drying agent, $H_{2} O v$ are absorbed
$P_{N_{2}}^{'} = 745 m m$
$P_{H_{2} O}^{'} = 760 - 745 = 15 m m$
$P_{N_{2}}^{'} = P_{M} \times$ mole fraction of $N_{2}$
mole fraction of $N_{2} = \frac{P_{N_{2}}}{P_{M}} = \frac{745}{760} = 0.9803$
% mole fraction of $N_{2} = 98.03$ %
% mole fraction of $H_{2} O = 1.97$ %
So, nearest integer value is 98.

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Given, P′N2+P′H2O=760mmAfter the use of drying agent, H2Ov are absorbedP′N2=745mmP′H2O=760−745=15mmP′N2=PM× mole fraction of N2mole fraction of N2=PN2PM=745760=0.9803% mole fraction of N2=98.03%% mole fraction of H2O=1.97%So, nearest integer value is 98.