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(C) Let the mass of both helium $(He)$ and oxygen $(O_2)$ be $m \ g$. The molar mass of $He$ is $4 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol

Vedclass · Accepted Answer

$\frac{8}{9}$

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(C) Let the mass of both helium $(He)$ and oxygen $(O_2)$ be $m \ g$. The molar mass of $He$ is $4 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$. The number of moles of $He$ $(n_{He})$ = $\frac{m}{4}$. The number of moles of $O_2$ $(n_{O_2})$ = $\frac{m}{32}$. The total number of moles $(n_{total})$ = $\frac{m}{4} + \frac{m}{32} = \frac{8m + m}{32} = \frac{9m}{32}$. According to Dalton's law of partial pressures,the fraction of total pressure exerted by a gas is equal to its mole fraction. Mole fraction of $He$ $(x_{He})$ = $\frac{n_{He}}{n_{total}} = \frac{m/4}{9m/32} = \frac{m}{4} 	imes \frac{32}{9m} = \frac{8}{9}$.

Equal masses of helium and oxygen are mixed in an empty container at $25^{\circ} C$. What is the fraction of the total pressure exerted by helium?

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