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 760mm. After some hours the pressure reached a sready value of 745mm.a) calculate the composition in mole% of the original mixture.b) if the experiments is done at 20°c and the drying agent increases in weight by 0•15g what is the volume of the flask.

T = 273.15 + 20 = 293.15 KP (N2) = P = 745 mm Hg P (Water) = 760 – 745 = 15 mm Hg Lets assume that the portion of  nitrogen is Xn, then as per the concept of partial pressure,P(N2) = Xn x P = 745 / 760 = 0.98Hence, the percentage of mole = 98%Mass of water vapor = Increases in weight of drying agent = 0.15gP(water) = 15 mm Hg  exerted by 0.15 g of water, then by formulaV = nRT/P= [(0.15/18) x (8.205 x 10-2 x 293.15)]/(15/760)= 10.2 L