Question

naturally occurring boron consists of two isotopes whose atomic masses are 10.01 and 11.01 . The atomic mass of natural boron is 10.81 . calculate the percentage of each isotope in natural boron .

Answer 1

Answer:

Let the percentage of boron isotope with atomic weights 10.01  be x% and that of atomic weight 11.01 be (100-x)%.  The average atomic mass of boron is 10.81.

Average atomic mass =  

100

The atomic weight of the first isotope×x+atomic weight of second isotope×(100−x)

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10.81=  100

10.01x+11.01×(100−x)

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x=20

(100−x)=80

Hence, the natural abundance of an isotope with atomic weight 10.01 is 20%  

the natural abundance of an isotope with atomic weight 11.01 is 80%

Answer 2

80% is the percentage of each isotope in natural boron.

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