A naturally occurring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81)

A. 20% and 80%

B. 20% and 80%

C. 25% and 75%

D. 75% and 25%

Solution: (A)

Let x% be the percentage of an isotope with atomic weights 10.01 and (100 – x)% be the percentage of an isotope with an atomic weight of 11.01

Average atomic weight = [(atomic weight of first isotope) x (x) + (atomic weight of second isotope) x (100-x)] / x + (100 – x)

10.81 = (x) x (10.01) + (100 – x) x (11.01) / x + (100 – x)

10.81 = [10.01x + 11.01 x (100 – x)] / 100

We get,

x = 20

Therefore,

Percentage of the isotope with atomic weight 10.01 = 20

Percentage of isotope with atomic weight 11.01 = 100 – 20 = 80

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