Mole fraction of O_2 in an O_2 and O_3 mixtur | vedclass

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Question

(A) Let the total number of moles be $n_{total} = 1 \ mol$. Given mole fraction $X_{O_2} = 0.25$,so $n_{O_2} = 0.25 \ mol$. Consequently,$n_{O_3} = 1

Vedclass · Accepted Answer

$18.18$

Vedclass · Answer

(A) Let the total number of moles be $n_{total} = 1 \ mol$. Given mole fraction $X_{O_2} = 0.25$,so $n_{O_2} = 0.25 \ mol$. Consequently,$n_{O_3} = 1 - 0.25 = 0.75 \ mol$. Mass of $O_2$ $(w_{O_2})$ = $n_{O_2} 	imes M_{O_2} = 0.25 \ mol 	imes 32 \ g/mol = 8 \ g$. Mass of $O_3$ $(w_{O_3})$ = $n_{O_3} 	imes M_{O_3} = 0.75 \ mol 	imes 48 \ g/mol = 36 \ g$. Percentage concentration $\left( \frac{w}{W}\% ight)$ of $O_2 = \frac{w_{O_2}}{w_{O_2} + w_{O_3}} 	imes 100 = \frac{8}{8 + 36} 	imes 100 = \frac{8}{44} 	imes 100 = \frac{2}{11} 	imes 100 \approx 18.18\%$.

Mole fraction of $O_2$ in an $O_2$ and $O_3$ mixture is $0.25$. Hence,the percentage concentration $\left( \frac{w}{W}\% \right)$ of $O_2$ in the mixture is:

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