Question

What will be the ratio of $C{l}^{35}$ and $C{l}^{37}$ respectively in chlorine if the average atomic mass of chlorine is 35.5?

• A. 1 : 1
• B. 3 : 1
• C. 2 : 1
• D. 3 : 2

Answer 1

Answer: (B)

Let percentage abundance of $C{l}_{37}=x$

Percentage abundance of $C{l}_{35.5}=1-x$

Average atomic mass $=\Sigma \left(mas{s}_{\left(i\right)}\right)\left(abundanc{e}_{\left(i\right)}\right)$

$37x+35(1-x)=35.5$

$37x+35-35x=35.5$

$2x=0.5$

$x=0.25$

Hence, 25% $C{l}_{37}$ and 75% $C{l}_{35}$.

So, the ratio of $C{l}_{35}$ and $C{l}_{37}$ is $3:1$.

Hence, option C is correct.