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Ammonia

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Question

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

A

443.34 L

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B

246.3 L

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C

12.315 L

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D

24.63 L

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Solution

The correct option is D
24.63 L

$P_{H_{2} O} = 760 - 722 = \frac{38}{760} a t m . n H_{2} O = \frac{0.9}{18} V$

$V = \frac{0.9}{18} \times \frac{0.821 \times 300}{38} \times 760; V = 24.63 L$

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