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Pressure

Same mass of ...

Question

Same mass of $C H_{4}$ and $H_{2}$ is taken in a container. The partial pressure caused by $H_{2}$ will be what fraction of total pressure?

A

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B

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C

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D

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Solution

The correct option is A
Partial pressure = mole fraction $\times$ total pressure
Mole fraction of $H_{2}$ = $\frac{\frac{m}{M_{C H_{4}}}}{\frac{m}{M_{C H_{4}}} + \frac{m}{M_{H_{2}}}} = \frac{\frac{1}{2}}{\frac{1}{16} + \frac{1}{2}} = \frac{8}{9}$

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