A mixture of equal masses of two gases with m | vedclass

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(D) Let the mass of each gas be $m \, g$. Number of moles of the lighter gas $(M_1 = 4)$ is $n_1 = \frac{m}{4}$. Number of moles of the heavier gas $(

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(D) Let the mass of each gas be $m \, g$. Number of moles of the lighter gas $(M_1 = 4)$ is $n_1 = \frac{m}{4}$. Number of moles of the heavier gas $(M_2 = 40)$ is $n_2 = \frac{m}{40}$. Total moles $n_{total} = n_1 + n_2 = \frac{m}{4} + \frac{m}{40} = \frac{10m + m}{40} = \frac{11m}{40}$. Mole fraction of the lighter gas $(x_1)$ = $\frac{n_1}{n_{total}} = \frac{m/4}{11m/40} = \frac{m}{4} 	imes \frac{40}{11m} = \frac{10}{11}$. Partial pressure of the lighter gas = $x_1 	imes P_{total} = \frac{10}{11} 	imes 1.1 \, atm = 1.0 \, atm$.

$A$ mixture of equal masses of two gases with molecular weights $4$ and $40$ has a total pressure of $1.1 \, atm$. The partial pressure of the lighter gas in the mixture is ............ $atm$.

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