Two isotopes of Chlorine  are :  \(^{35}\)Cl  and  \(^{37}\)Cl 

 Atomic mass of Cl :  35.5

Ratio of the isotopes of Chlorine : 3 : 1

Let the relative atomic abundance of \(^{37}\)Cl  =  1 x X =  X

Let the relative atomic abundance of \(^{35}\)Cl  = 3 x X = 3X

N.B : The average atomic mass = sum of products of each isotope mass and its relative abundance.

  A.M.M = (3x of 35)  +  (x of 37)

  35.5 = (3x X 35)  + (x  X 37)

  35.5 = 105x  + 37x

  35.5 = 142x

 x = \(\frac{35.5}{142}\)

 x =  0.25

Convert to percentage.

Recall that relative atomic abundance of \(^{37}\)Cl  =  x = 0.25 x  100 = 25%

 and the relative atomic abundance of \(^{35}\)Cl  = 3X = 3(0.25) = 0.75 x 100 = 75%
We were asked for the Relative abundance of isotope of mass number 37 i.e \(^{37}\)Cl  is 25%

Alternatively,

Ratio of the isotopes of Chlorine : 3 :  1 

Sum of ratios = 3 + 1 = 4

Relative Abundance of Cl isotopes = % ratio of \(^{35}\)Cl  and  % ratio of  \(^{37}\)Cl 

  = \(\frac{3}{4}\)x 100  and \(\frac{1}{4}\) x 100

  = 75% \(^{35}\)Cl  and  25% \(^{37}\)Cl  

Relative abundance of \(^{37}\)Cl = 25