Equal weights of methane and oxygen are mixed | vedclass

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(A) Let the weight of both methane $(CH_4)$ and oxygen $(O_2)$ be $32\,g$.Molar mass of $CH_4 = 16\,g/mol$.Molar mass of $O_2 = 32\,g/mol$.Moles of $C

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$\frac{1}{3}$

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(A) Let the weight of both methane $(CH_4)$ and oxygen $(O_2)$ be $32\,g$.Molar mass of $CH_4 = 16\,g/mol$.Molar mass of $O_2 = 32\,g/mol$.Moles of $CH_4$ $(n_{CH_4})$ = $\frac{32\,g}{16\,g/mol} = 2\,mol$.Moles of $O_2$ $(n_{O_2})$ = $\frac{32\,g}{32\,g/mol} = 1\,mol$.Total moles $(n_T)$ = $2 + 1 = 3\,mol$.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 $(P_T)$.Mole fraction of $O_2$ $(X_{O_2})$ = $\frac{n_{O_2}}{n_T} = \frac{1}{3}$.Therefore,the fraction of the total pressure exerted by oxygen is $X_{O_2} = \frac{1}{3}$.

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

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