Question

For every ,one ${}^{37}Cl$ isotope there are three ${}^{35}Cl$ isotopes, in a sample of chlorine. What will be the average atomic mass of chlorine?

• A. 35
• B. 37
• C. 35.5
• D. 35.6

Answer 1

The correct option is C 35.5

Average atomic mass of an element existing in different isotopes is given by:

$M_{avg}=\frac{{\sum }_{i=1}^n{M}_i{A}_i}{{\sum }_{i=1}^n{A}_i}$

where $M_i=$atomic mass of an isotope with relative abundance of $A_i$

Given:$M_1=37u,A_1=1,M_2=35u,A_2=3$

on subtitutiing we get:

$M_{avg}=\frac{37\times 1+35\times 3}{1+3}$

$M_{avg}=35.5u$

option C is correct