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Question

Boron has 2 isotopes B-10 B-11. Th average atomic mass of boron is found to be 10.80u. Calculate the percentage abundance of these isotopes.

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Solution

Given Average Atomic Mass of B = 10.80 amu
Let the percentage of 10B isotope = x
Then the percentage of 11B isotope = 100 - x

Average Atomic Mass Of B = X×10 +(100 - x)×11100
Thus
X×10 +(100 - x)×11100 = 10.80 10x + 1100 - 11x = 10.8 × 100 1100-x = 1080 -x = 1080-1100 -x = -20 x =20

Thus percentage abundance of 10B = 20 %
and percentage abundance of 11B = 100 - 20 % = 80 %


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