Equal weights of CH_{4} and H_{2} are mixed i | vedclass

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(C) Let the weight of each gas be $x \ g$. Number of moles of $CH_{4} = \frac{x}{16} \ mol$. Number of moles of $H_{2} = \frac{x}{2} \ mol$. Total mol

Vedclass · Accepted Answer

$8 / 9$

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(C) Let the weight of each gas be $x \ g$. Number of moles of $CH_{4} = \frac{x}{16} \ mol$. Number of moles of $H_{2} = \frac{x}{2} \ mol$. Total moles = $\frac{x}{16} + \frac{x}{2} = \frac{x + 8x}{16} = \frac{9x}{16} \ mol$. The mole fraction of $H_{2}$ is $X_{H_{2}} = \frac{n_{H_{2}}}{n_{total}} = \frac{x/2}{9x/16} = \frac{x}{2} 	imes \frac{16}{9x} = \frac{8}{9}$. According to Dalton's Law, the partial pressure is proportional to the mole fraction. Therefore, the fraction of total pressure exerted by $H_{2} = \frac{8}{9}$.

Equal weights of $CH_{4}$ and $H_{2}$ are mixed in an empty container at $25^{\circ} C$. The fraction of the total pressure exerted by $H_{2}$ is

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