Question

3-6.
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.
(a) Calculate the composition, in mol and per cent of 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 1

Answer:

T=273.15+293.15KP(N2)

P=745mm 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*P

=745/760

=0.98

Hence the percentage of mole =98% mass of water vapor=Increases in weight of drying agent=0.15g

P(water)=15 mm Hg is exerted by 0.15 g of water so finally

V=nRT/P

=[(0.15/18)*(8.205*10-2*293.15)]/(15/760)

=10.2L