Question

Let the percentage of relative abundance of the two isotopes of chlorine be ( x ) %and ( (100-x) % ) Atomic mass of chlorine= ( (x / 100 times 35)+(100-x / 100 times 37) )
( 35.5=(35 times / 100)+(3700-37 times 1100) )
( 35.5=(35 x+3700-37 x) / 100 )
( 35.5=3700-2 times / 100 )
( 3550=3700-2 x )
( -2 x=3550-3700 )
( -2 x=-150 Rightarrow x=75 )
Percentage of relative abundance of ( mathrm{Cl}_{35}=75 % ) Percentage of relative abundance of ( mathrm{Cl}_{37}=100-75=15 % )

# Chleune exists as 2 isotopes 35 and 37 with a average atemic mass of 35.5 grams. Then determine the sef abundance of e 35 isotope.

Solution