Equal masses of CH_4 and H_2 are mixed in an | vedclass

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(A) Let the mass of both $CH_4$ and $H_2$ be $x \ g$. Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{x}{16} \ mol$. Number of moles of $H_2$ $(n_{H_2

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$1/9$

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(A) Let the mass of both $CH_4$ and $H_2$ be $x \ g$. Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{x}{16} \ mol$. Number of moles of $H_2$ $(n_{H_2})$ = $\frac{x}{2} \ mol$. The mole fraction of $CH_4$ $(X_{CH_4})$ is given by $\frac{n_{CH_4}}{n_{CH_4} + n_{H_2}}$. $X_{CH_4} = \frac{x/16}{x/16 + x/2} = \frac{1/16}{1/16 + 8/16} = \frac{1/16}{9/16} = 1/9$. According to Dalton's Law of partial pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure $(P_{CH_4} = X_{CH_4} 	imes P_{Total})$. Therefore,the fraction of the total pressure exerted by methane is $1/9$.

Equal masses of $CH_4$ and $H_2$ are mixed in an empty container at $25\,^oC$. The fraction of the total pressure exerted by methane is

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