A mixture of H_2 and O_2 at 1 , atm pressure | vedclass

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(D) Let the total weight of the mixture be $100 \, g$.Weight of $H_2 = 20 \, g$ and weight of $O_2 = 80 \, g$.Moles of $H_2$ $(n_{H_2})$ = $\frac{20 \

Vedclass · Accepted Answer

$0.8$

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(D) Let the total weight of the mixture be $100 \, g$.Weight of $H_2 = 20 \, g$ and weight of $O_2 = 80 \, g$.Moles of $H_2$ $(n_{H_2})$ = $\frac{20 \, g}{2 \, g/mol} = 10 \, mol$.Moles of $O_2$ $(n_{O_2})$ = $\frac{80 \, g}{32 \, g/mol} = 2.5 \, mol$.Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{10}{10 + 2.5} = \frac{10}{12.5} = 0.8$.Partial pressure of $H_2$ $(P_{H_2})$ = $x_{H_2} 	imes P_{Total} = 0.8 	imes 1 \, atm = 0.8 \, atm$.

$A$ mixture of $H_2$ and $O_2$ at $1 \, atm$ pressure contains $20 \%$ by weight of $H_2$. Find out the partial pressure of $H_2$ in $atm$.

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