Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron.

Let x% be the percentage of isotope with atomic weights 10.01 and

(100 – x)% be the percentage of isotope with atomic weight 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 isotope with atomic weight 10.01 = 20

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

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