A mixture of nitrogen 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 some hours the pressure reached a steady value of 745 mm.
Text Solution
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1.98% 10.156 litres
By Dalton's partial pressure
+ 
= 760 mm
From given data 
= 745 mm
So 
= 760 – 745 = 15 mm
% Mole of N 2 = % of pressure of N 2 = 
× 100 = 98.02
∴ Mole % of H 2 O = 100 – 98.02 = 1.98% Ans.
Increase weight of drying agent due to absorption of water (H 2 O).
Hence, Wt. of H 2 O = 0.15 g
∴ Mole of H 2 O = 
Pressure of H 2 O( 
) = 15 mm = 
atm
From gas equation PV = nRT
× V = 
× 0.0821 × (273 + 20)
V = 10.156 litres Ans.
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