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Volume of Gases and Number of Moles

A gaseous mix...

Question

A gaseous mixture of $O_{2}$ and $N_{2}$ in a closed vessel having mass 1g occupies $750 m L$ at NTP. Volume of $O_{2}$ in the mixture at $N T P$ is?

A

400 m L

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B

350 m L

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C

450 m L

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D

300 m L

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Solution

The correct option is C $350 m L$
We know that
No. of moles $= \frac{v o l u m e o f s o l u t i o n}{v o l u m e o f s o l u t i o n a t N T P}$
No. of moles of $O_{2}$ + No. of moles of $N_{2}$
Let mass of $N_{2}$ is $x$
So, mass of $O_{2}$ is $1 - x$
Total moles $= \frac{1 - x}{32} + x 28$
So, on putting the values in formula, we have
$\frac{1 - x}{32} + x 28 = \frac{750 L}{22.4 L * 1000}$
On solving
$x = 0.5$
So, mass of $O_{2} = 1 - 0.5 = 0.5$ g
Volume of $O_{2}$ in solution
No. of moles $O_{2} = \frac{v o l u m e o f O_{2} i n s o l u t i o n}{22.4 L}$
$ \frac {0.5} {32} = \frac {V} {22.4 L}
$V = 350 m l$

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