Equal masses of methane and oxygen are mixed | vedclass

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(D) Let the mass of methane $(CH_4)$ and oxygen $(O_2)$ be $m \ g$.Moles of $CH_4 = \frac{m}{16}$Moles of $O_2 = \frac{m}{32}$According to Dalton's La

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$1/3$

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(D) Let the mass of methane $(CH_4)$ and oxygen $(O_2)$ be $m \ g$.Moles of $CH_4 = \frac{m}{16}$Moles of $O_2 = \frac{m}{32}$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.Mole fraction of $O_2 (x_{O_2}) = \frac{	ext{Moles of } O_2}{	ext{Moles of } O_2 + 	ext{Moles of } CH_4}$$x_{O_2} = \frac{m/32}{m/32 + m/16} = \frac{m/32}{m/32 + 2m/32} = \frac{m/32}{3m/32} = 1/3$Therefore,the fraction of the total pressure exerted by oxygen is $1/3$.

Equal masses 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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