Question

At ${1200}^{o}C$, mixture of $C{l}_2$ and $Cl$ atoms (both in gaseous state) effuses 1.16 times faster than krypton effuses under identical conditions. The fraction of chlorine molecules dissociated into atoms is approximate:

[MW of Kr $=$ 84]

• A. 13%
• B. 22%
• C. 87%
• D. 44%

Answer 1

Answer: (A)

13%

The rate of effusion of a gas is inversely proportional to its molecular weight.

$r\propto \frac{1}{\sqrt{M}}$.

Let $x$ be the average molecular weight of a mixture of $C{l}_2$ and $Cl$ atoms.The molecular weight of krypton is 84.

$\frac{r_{\text{(mixture of}C{l}_2\text{and Cl atoms)}}}{r_{\text{argon}}}=\sqrt{\frac{M_{\text{argon}}}{M_{\text{(mixture of}C{l}_2\text{and Cl atoms)}}}}$.

Substitute values in the above expression, we get:

$1.16=\sqrt{\frac{84}{x}}$ or $x=62.43$

Also, $62.43=\frac{(100-y)\times 71+y\times 35.5}{100}$ or $y=24.1$.

So, in 1000 moles species 759 moles is $C{l}_2$ molecules and 241 moles is $Cl$ atom.

So, initial moles of $C{l}_2=759+\frac{241}{2}$

So, % fraction dissociated $\frac{\frac{241}{2}}{759+\frac{241}{2}}\times 100=13.7\%$

Hence, the correct option is $\text{A}$