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Question
Solve the problem:
The natural isotopic abundance of 10B is 19.60% and 11B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
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Solution
| Isotope | % Abundance | Mass (u) |
| ¹⁰B | 19.60% | 10.13 u |
| ¹¹B | 80.40% | 11.009 u |
Convert % to fractional abundance
For ¹⁰B: 19.60% = 0.1960
For ¹¹B: 80.40% = 0.8040
Apply formula for average atomic mass
Average atomic mass = (f1 × m1) + (f2 × m2)
= (0.1960 × 10.13) + (0.8040 × 11.009)
0.1960 × 10.13 = 1.98648
0.8040 × 11.009 = 8.84724
1.98648 + 8.84724 = 10.83372
Average atomic mass of boron = 10.83u (approx.)
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