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Hydrogen is collected over water at 22° C and a barometer reading of 7 40 mm.Hg. If 300 mL of hydrogen is collected, which expression will give the volume of dry hydrogen at the same temperature and pressure?
Compound Vapour Pressure at 22° C
$H_{2} O$ 20 mm Hg

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$300 mL \times \frac{740 mmHg - 20 mmHg}{740 mmHg}$

$300 mL \times \frac{740 mmHg + 20 mmHg}{740 mmHg}$

$300 mL \times \frac{740 mmHg}{740 mmHg - 20 mmHg}$

$300 mL \times \frac{740 mmHg}{740 mmHg + 20 mmHg}$

answer is A.

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Detailed Solution

Volume of dry hydrogen at the same temperature and pressure;

$V = 300 \times \frac{700 - 200}{740}$

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