A mixture of dihydrogen and dioxygen at 1 ba | vedclass

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(A) Let the weight of dihydrogen be $20 \ g$ and the weight of dioxygen be $80 \ g$.The number of moles of dihydrogen is $n_{H_2} = \frac{20 \ g}{2 \

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$0.8$

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(A) Let the weight of dihydrogen be $20 \ g$ and the weight of dioxygen be $80 \ g$.The number of moles of dihydrogen is $n_{H_2} = \frac{20 \ g}{2 \ g/mol} = 10 \ mol$.The number of moles of dioxygen is $n_{O_2} = \frac{80 \ g}{32 \ g/mol} = 2.5 \ mol$.The total pressure of the mixture is $P_{total} = 1 \ bar$.The partial pressure of dihydrogen is given by $p_{H_2} = \chi_{H_2} 	imes P_{total}$,where $\chi_{H_2}$ is the mole fraction of dihydrogen.$\chi_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{10}{10 + 2.5} = \frac{10}{12.5} = 0.8$.Therefore,$p_{H_2} = 0.8 	imes 1 \ bar = 0.8 \ bar$.

$A$ mixture of dihydrogen and dioxygen at $1 \ bar$ pressure contains $20 \%$ by weight of dihydrogen. Calculate the partial pressure of dihydrogen. (in $bar$)

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