KVPY SX SB Exam 05-Nov-2017 Question Paper

$\mathit{M}\mathit{a}\mathit{v}=\frac{{\mathit{M}}_1{\mathit{n}}_1+{\mathit{M}}_2{\mathit{n}}_2}{{\mathit{n}}_1+{\mathit{n}}_2}$
$35.45=\frac{35{\mathit{n}}_1+37{\mathit{n}}_2}{{\mathit{n}}_1+{\mathit{n}}_2}$
∴ ${\mathit{n}}_1:{\mathit{n}}_2=3:1$


(OR)



The percent abundance of $\mathit{C}\mathit{l}-35$ and $\mathit{C}\mathit{l}-37$ is equal to 100 .
Let abundance of ${}^{35}\mathit{C}\mathit{l}=\mathit{x}$ and abundance of ${}^{37}\mathit{C}\mathit{l}=100-\mathit{x}$.
As we know, Atomic weight of ${}^{35}\mathit{C}\mathit{l}=34.9689$
Atomic weight of ${}^{37}\mathit{C}\mathit{l}=36.9695$
Atomic mass of $\mathit{C}\mathit{l}=35.45$
Average atomic mass $=\frac{34.9689\left(\mathit{x}\right)+36.9695\left(100-\mathit{x}\right)}{100}=35.45$
$\mathit{x}=76\%$
Abundance of $\mathit{C}\mathit{l}-35=76\%$
Abundance of $\mathit{C}\mathit{l}-37=24\%$
Hence, the ratio of their natural abundance is approximately equal to $3:1.$