At STP, a container has 1 mole of Ar, 2 moles of ${CO}_{2}$, 3 moles of $O_{2}$ and 4 moles of $N_{2}$. Without changing the total pressure if one mole of $O_{2}$ is removed, the partial pressure of $O_{2}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Mole fraction of A =$\frac{moles\ of\ A\ }{total\ moles\ }$
Given,
moles of Ar = 1, moles of ${CO}_{2}$ = 2,
moles of $O_{2}$ = 3, moles of $N_{2}$ = 4,
moles of $O_{2}$ removed = 1
Mole fraction of $O_{2}$ at initial stage
=$\frac{3}{1 + 2 + 3 + 4} 	imes$100=$\frac{3}{10} 	imes 100 = 30$
Mole fraction of $O_{2}$ at final stage
= $\left( \frac{3}{10} - \frac{2}{1 + 2 + 2 + 4} ight) 	imes 100\ \$
=$\left( \frac{3}{10} - \frac{2}{9} ight) 	imes 100 = \frac{70}{9}\  \cong \ 8$
$	herefore$ % change $= \ \frac{8}{30}\  	imes \ 100\  = 26\%$

Q. At STP, a container has 1 mole of Ar, 2 moles of ${CO}_{2}$, 3 moles of $O_{2}$ and 4 moles of $N_{2}$. Without changing the total pressure if one mole of $O_{2}$ is removed, the partial pressure of $O_{2}$ is

Explanation

Reviewed by Dr. Abubacker

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