Mole fraction of O_2 in O_2 and O_3 mixture i | 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 $n_{O_2} = 0.25 \ mol$.Then $n_{O_3} = 1 - 0.25 = 0

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 $n_{O_2} = 0.25 \ mol$.Then $n_{O_3} = 1 - 0.25 = 0.75 \ mol$.Mass of $O_2 = n_{O_2} 	imes M_{O_2} = 0.25 \ mol 	imes 32 \ g/mol = 8 \ g$.Mass of $O_3 = n_{O_3} 	imes M_{O_3} = 0.75 \ mol 	imes 48 \ g/mol = 36 \ g$.Total mass of mixture $= 8 \ g + 36 \ g = 44 \ g$.Percentage concentration $\left( \frac{w}{W} \% ight)$ of $O_2 = \frac{	ext{mass of } O_2}{	ext{total mass}} 	imes 100 = \frac{8}{44} 	imes 100 = \frac{2}{11} 	imes 100 \approx 18.18 \%$.

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

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