Let x = fraction of Cl-35
Let y = fraction of Cl-37
x + y = 1
So, y = 1 - x
Given, atomic mass of Cl-35 = 34.96; atomic mass ofCl-37 = 36.95; average mass of chlorine = 35.43
34.96 × x + 36.95 × y = 35.43
34.96 × x + 36.95 × (1 - x) = 35.43 (since, y = 1 - x)
Solving for x gives x = 0.7595
Solving for y gives y = 0.2405
Multiplying with 100 gives percentage abundance:
Percentage abundance of Cl-35 = 0.7595 × 100 = 75%
Percentage abundance of Cl-37 = 0.2405 × 100 = 24.05%