A container contains a mixture of He and Ne g | vedclass

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(A) Let the total mass of the mixture be $100 	ext{ g}$.Mass of $He$ $(m_{He})$ = $20 	ext{ g}$, Mass of $Ne$ $(m_{Ne})$ = $80 	ext{ g}$.Molar mass

Vedclass · Accepted Answer

$5:4$

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(A) Let the total mass of the mixture be $100 	ext{ g}$.Mass of $He$ $(m_{He})$ = $20 	ext{ g}$, Mass of $Ne$ $(m_{Ne})$ = $80 	ext{ g}$.Molar mass of $He$ $(M_{He})$ = $4 	ext{ g/mol}$, Molar mass of $Ne$ $(M_{Ne})$ = $20 	ext{ g/mol}$.Number of moles of $He$ $(n_{He})$ = $\frac{20 	ext{ g}}{4 	ext{ g/mol}} = 5 	ext{ mol}$.Number of moles of $Ne$ $(n_{Ne})$ = $\frac{80 	ext{ g}}{20 	ext{ g/mol}} = 4 	ext{ mol}$.According to Dalton's Law, the ratio of partial pressures is equal to the ratio of the number of moles: $\frac{P_{He}}{P_{Ne}} = \frac{n_{He}}{n_{Ne}} = \frac{5}{4}$.

$A$ container contains a mixture of $He$ and $Ne$ gases at a certain temperature. $He$ is $20 \%$ by mass of the mixture. What is the ratio of the partial pressure of $He$ to $Ne$?

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