The relative atomic mass of boron is 10 . . calculate the percentage of its isotopes 105B and 115 B occurring in nature
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Shweta Gulati 8 years, 3 months ago
Suppose percentage of isotope 105B = x
Then , percentage of isotope 115B = (100-x)
Average atomic mass = [10 X x + 11(100-x) ]/100
= (1100-x)/100
This is equal to 10.81 (Given) [ This is the average atomic mass of Boron]
1100 -x =1081
1100 -1081 = x
x = 19
Hence, % of 105B = 19%
and % of 115B = (100 -19) = 81%
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