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(D) Let $n_1$ be the number of moles of hydrogen $(H_2)$ and $n_2$ be the number of moles of helium $(He)$.Using the ideal gas equation $PV = nRT$,whe

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$2: 5$

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(D) Let $n_1$ be the number of moles of hydrogen $(H_2)$ and $n_2$ be the number of moles of helium $(He)$.Using the ideal gas equation $PV = nRT$,where $n = n_1 + n_2$ is the total number of moles.Given: $P = 2 \ atm = 2 	imes 1.013 	imes 10^5 \ Pa$,$V = 20 \ L = 20 	imes 10^{-3} \ m^3$,$T = 27^{\circ} C = 300 \ K$,$R = 8.314 \ J/mol \cdot K$.Total moles $n = n_1 + n_2 = \frac{PV}{RT} = \frac{2 	imes 1.013 	imes 10^5 	imes 20 	imes 10^{-3}}{8.314 	imes 300} \approx 1.62 \ mol$.Mass of mixture: $m_{H_2} + m_{He} = 5 \ g$.Since $m = n 	imes M$,where $M$ is molar mass: $n_1(2) + n_2(4) = 5 \implies n_1 + 2n_2 = 2.5$.Subtracting the first equation $(n_1 + n_2 = 1.62)$ from the second $(n_1 + 2n_2 = 2.5)$: $n_2 = 0.88 \ mol$.Then $n_1 = 1.62 - 0.88 = 0.74 \ mol$.Mass ratio $\frac{m_{H_2}}{m_{He}} = \frac{n_1 	imes 2}{n_2 	imes 4} = \frac{0.74 	imes 2}{0.88 	imes 4} = \frac{1.48}{3.52} = \frac{2}{5}$.

$A$ vessel of volume $20 \ L$ contains a mixture of hydrogen and helium at a temperature of $27^{\circ} C$ and pressure $2 \ atm$. The mass of the mixture is $5 \ g$. Assuming the gases to be ideal,the ratio of the mass of hydrogen to that of helium in the given mixture will be:

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