Question

Chlorine exists in two isotopic forms, $Cl-37$ and $Cl-35$ but its atomic mass is $35.5$. This indicates the ratio of $Cl-37$ and $Cl-35$ is approximately:

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

Answer 1

Answer: (C)

Let relative abundance of Cl-37 be $A_1$% with mass $M_1=37$ amu

$\therefore$ The relative abundance of Cl-35 = $A_2$% with mass, $M_2=35$ amu

where $A_1+A_2=100$

Thus $A_2$%=$(100-A_1)$%

Using average atomic mass, $M_{avg}=\frac{M_1{A}_1+M_2{A}_2}{A_1+A_2}$

$M_{avg}=\frac{37\times A_1+35\times (100-A_1)}{A_1+(100-A_1)}=\frac{2A_1+3500}{100}$

Given that: $M_{avg}=35.5$

$\therefore \frac{2A_1+3500}{100}=35.5$

$2A_1=35.5\times 100-3500=50$

$A_1=\frac{50}{2}=25$ and $A_2=100-A_1=100-25$ =75

Thus the ratio of isotope abundance of $Cl-37toCl-35$ is $A_1:A_2=25:75=1:3$

So, the correct option is $C$