Question

A mixture of nitrogen and water vapours is admitted to a flask at $760torr$ which contains a sufficient solid drying agent after long time the pressure reached a steady value of $722torr$. If the experiment is done at ${27}^{\circ }C$ and drying agent increases in weight by $0.9gm$, what is the volume of the flask? Neglect any possible vapour pressure of drying agent and volume occupied by drying agent.

• A. $443.34L$
• B. $246.3L$
• C. $12.315L$
• D. $24.63L$

Answer 1

The correct option is D $24.63L$

$P_{H_{2}O}=P_{Total}-P_{N_2}$

Given $P_{Total}$ = 760 torr, $P_{N_2}$ = 722 torr, T=300K

$P_{H_{2}O}=760-722=38$ torr

$P_{H_{2}O}=\frac{38}{760}$ atm

$n_{H_{2}O}=\frac{0.9}{18}$

Now,

$PV=nRT$

$V=\frac{nRT}{P}$

$V=\frac{0.9\times 0.821\times 300\times 760}{18\times 38}$

$V=24.63$ L

Hence the correct option is [D]