Chlorine has two stable isotopes, cl-35 and cl-37. if their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is the natural abundance of cl-35? (the atomic mass of chlorine is 35.45 amu)a) 75.95%b) 24.05%c) 50.00%d) 35.00%e) 37.00%

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

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