Calculate the percentage of the naturally occuring isotopes `""^(35)Cl` and `""^(37)Cl` that accounts for the atomic mass of chlorine taken as 35.45.

Explanation

Suppose `""^(37)Cl` present = `x%`. Then `""^(37)Cl` present =(100 -x)%
`therefore` Average atomic mass `=(x xx 35 + (100 -x) xx 37)/(100) = 35. 45` (Given)
or `35x + 3700 - 37 x = 3545` or `2x = 155` or `x = 77.5%`
This `""^(35)Cl = 77.5%` and `""^(37)Cl = (100 - 77.5) = 22.5` %