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 .
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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)
10.81= 100
10.01x+11.01×(100−x)
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%
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80% is the percentage of each isotope in natural boron.
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