x g of molecular oxygen (O_2) is mixed with | vedclass

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(B) According to Dalton's Law of Partial Pressures,$P_{total} = P_{O_2} + P_{Ne}$.Given $P_{total} = 25 \ bar$ and $P_{Ne} = 20 \ bar$,so $P_{O_2} = 2

Vedclass · Accepted Answer

$80$

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(B) According to Dalton's Law of Partial Pressures,$P_{total} = P_{O_2} + P_{Ne}$.Given $P_{total} = 25 \ bar$ and $P_{Ne} = 20 \ bar$,so $P_{O_2} = 25 - 20 = 5 \ bar$.Partial pressure is related to mole fraction $(X)$ by $P_i = X_i 	imes P_{total}$.Moles of $Ne = \frac{200 \ g}{20 \ g \ mol^{-1}} = 10 \ mol$.Moles of $O_2 = \frac{x \ g}{32 \ g \ mol^{-1}} = \frac{x}{32} \ mol$.Mole fraction of $O_2$ is $X_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{x/32}{x/32 + 10}$.Using $P_{O_2} = X_{O_2} 	imes P_{total}$,we get $5 = \left( \frac{x/32}{x/32 + 10} ight) 	imes 25$.Dividing by $5$,we get $1 = \left( \frac{x/32}{x/32 + 10} ight) 	imes 5$.$1 = \frac{5x/32}{(x + 320)/32} = \frac{5x}{x + 320}$.$x + 320 = 5x \implies 4x = 320 \implies x = 80 \ g$.

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$A$ mixture of $0.5 \text{ mol}$ $N_2$ gas, $1.0 \text{ mol}$ $O_2$ gas and $1.5 \text{ mol}$ $H_2$ gas exerts a total pressure of $18 \text{ bar}$. Find the partial pressure of $O_2(g)$. (in $\text{ bar}$)

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