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.)

A mixture of nitrogen and water vapour is admitted to a flask which co

1.	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.)
Answer» Solution :(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.