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A 225 mL sample of $H_{2}$ is collected over water at 25^∘ C and 735 mm Hg pressure. Which expression. represents the set-up to find the volume of dry $H_{2}$ at $0^{\circ} C$ and 1 atmosphere?
[ VP of $H_{2} O$ at $25^{\circ} C = 24 m m$ of Hg]

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$V = \frac{225 \times (735 - 24) \times 273}{760 \times 298}$

$V = \frac{225 \times 760 \times 298}{(735 - 24) \times 273}$

$V = \frac{225 \times 273 \times 760}{(735 + 24) \times 298}$

$V = \frac{225 \times (735 + 24) \times 298}{760 \times 273}$

answer is A.

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Detailed Solution

It is known that,

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

By putting the values we get,

$V = \frac{225 \times (735 - 24) \times 273}{760 \times 298}$

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