Question

If a mixture of $O_2$ and $O_3$ is having $V_{rms}=12.92m/s$ at 300K. Calculate mole percentage of $O_2$ in the mixture.

• A. 35 %
• B. 20 %
• C. 66.67 %
• D. none of these

Answer 1

The correct option is B 20 %

Let the mole fraction of $O_2$ b3 $x$

So, $O_3=1-x$

$V_{rns}=\sqrt{\frac{3RT}{M}}$

$\Rightarrow 12.92=\sqrt{\frac{3\times 8.314\times 300}{M}}$

$M=\frac{3\times 8.314\times 300}{(12.42{)}^2}$

$=\frac{3\times 8.314\times 300}{(12.92{)}^2}$

$M=44.83$

$32x+(1-x)\times 48=44.83$

$\Rightarrow 32x+48-48x=44.83$

$16x=49-44-83$

$=3.17$

$x=\frac{3.17}{16}$

$=0.198$

$\cong 0.2$