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2 liters of moist hydrogen were collected over water at $26^{\circ} C$ at a total pressure of one atmosphere. On analysis, it was found that the quantity of $H_{2}$ collected was $0.0788$ mole. What is the mole fraction of $H_{2}$ in the moist gas?

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$0.989$

$0.897$

$0.953$

$0.967$

answer is D.

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Detailed Solution

The mol fraction of $H_{2}$ in the moist gas is,

$n = \frac{1 \times 2}{0.0821 \times 299}$

$= 0.0815$

$X_{H_{2}} = \frac{0.0788}{0.0815}$

$= 0.967$

Hence, option D is correct.

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