If equal masses of methane (CH_4) and hydroge | vedclass

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(D) Let the mass of both $CH_4$ and $H_2$ be $m \, g$.Number of moles of $CH_4$ $(n_{CH_4})$ = $m / 16$.Number of moles of $H_2$ $(n_{H_2})$ = $m / 2$

Vedclass · Accepted Answer

$8/9$

Vedclass · Answer

(D) Let the mass of both $CH_4$ and $H_2$ be $m \, g$.Number of moles of $CH_4$ $(n_{CH_4})$ = $m / 16$.Number of moles of $H_2$ $(n_{H_2})$ = $m / 2$.Total moles $(n_{total})$ = $n_{CH_4} + n_{H_2} = m/16 + m/2 = (m + 8m) / 16 = 9m / 16$.The partial pressure fraction of a gas is equal to its mole fraction.Mole fraction of $H_2$ $(x_{H_2})$ = $n_{H_2} / n_{total} = (m/2) / (9m/16) = (m/2) 	imes (16/9m) = 8/9$.Therefore,the fraction of total pressure exerted by hydrogen is $8/9$.

If equal masses of methane $(CH_4)$ and hydrogen $(H_2)$ are mixed in an empty container at $25\,^oC$,what is the fraction of the total pressure exerted by hydrogen?

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