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Chlorine consisting of two isotopes of mass number 35 and 37 in the ratio 3:1...

Chemistry
JAMB 2004

Chlorine consisting of two isotopes of mass number 35 and 37 in the ratio 3:1 has an atomic mass of 35.5. Calculate the relative abundance of the isotopes of mass number 37

  • A. 20
  • B. 25
  • C. 60
  • D. 75
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Correct Answer: Option B
Explanation

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

There is an explanation video available below.


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