A mixture of nitrogen and water vapour is admitted to a flask which contains a solid drying agent. Immediately after admission, the pressure of the flask is 760 mm. After some hours the pressure reached a steady value of 745 mm.
(a) Calculate the composition, in mole per cent, of the original mixture.
(b) If the experiment is done at `20^(@)C` and the drying agent increases in weight by 0.15 g, what is the volume of the flask ? (The volume occupied by the drying agent may be ignored.)

Explanation

(a) `p_(N_(2)) + p_(H_(2)O) = 760 mm`
`p_(N_(2)) = 745 mm`
`p_(H_(2)O) = 760-745 = 15 mm`.
Mole % of `N_(2)` = pressure % of `N_(2)`
`= (745)/(760) xx 100 = 98.03%`.
`therefore` mole % of `H_(2)O = 100 - 98.03 = 1.975%`.
(b) Since the weight of drying agent increases by 0.15 g,
weight of `H_(2)O = 0.15 g`.
Mol. wt of `H_(2)O = 18`.
`therefore` mole of `H_(2)O = (0.15)/(18)`
`p_(H_(2)O) = 15 mm = (15)/(760)` atm.
Applying `pV = nRT` in the flask of volume V litres (say)
`(15)/(760) xx V = (0.15) xx 0.082 (273 + 20)`
`V = 10.28` litres.