A bulb contains 1 g of H_2 and 1 g of He. I | vedclass

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(B) The number of moles of $H_2$ is $n_{H_2} = \frac{1 \ g}{2 \ g/mol} = 0.5 \ mol$.The number of moles of $He$ is $n_{He} = \frac{1 \ g}{4 \ g/mol} =

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The number of moles of $H_2$ is $n_{H_2} = \frac{1 \ g}{2 \ g/mol} = 0.5 \ mol$.The number of moles of $He$ is $n_{He} = \frac{1 \ g}{4 \ g/mol} = 0.25 \ mol$.The total number of moles is $n_{total} = 0.5 + 0.25 = 0.75 \ mol$.The mole fraction of $He$ is $x_{He} = \frac{n_{He}}{n_{total}} = \frac{0.25}{0.75} = \frac{1}{3}$.The partial pressure of $He$ is $P_{He} = x_{He} 	imes P_{total} = \frac{1}{3} 	imes 1.2 	imes 10^3 \ Pa = 0.4 	imes 10^3 \ Pa = 400 \ Pa$.

$A$ bulb contains $1 \ g$ of $H_2$ and $1 \ g$ of $He$. If the total pressure is $1.2 \times 10^3 \ Pa$,then the partial pressure of $He$ is ............. $Pa$.

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