Correct Answer - C
Let relative abundance of Cl-37 = x%
then relative abundance of Cl-35 = (100 - x)%
Average atomic mass `=(x xx37+(100-x)35)/100=35.5
rArr37x+3500-35x=3550rArrx=25
therefore" "100-x=75`
Thus, the ratio of Cl-37 and Cl-35 is x : (100 - x)
= 25 : 75 = 1 : 3