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Nuturally occurring boron consists of two isotopes whese atomic weight are `10.01` and `11.01`The atomic weight of nurons is `10.81` Calculate the percentage of each isotope is natural boron |
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Answer» Let the percentage of isotope with atomic weight `10.01` be x . So the percentage of isotope with atomic weight `11.01` will be `100 - x`. Average atomic weight = `(m_(1)x_(1) + m_(2) x_(2))/(s_(1) + s_(2))` `= (s xx 10.01 + (100 - s) xx 11.01)/(100)` `10.81 = (s xx 10.01 + (100 - s) xx 11.01)/(100) = 20` `%` of isotope with atomic weight `10.01 = 20` `%` of isotope with atomic weight `11.01 = 100 - x = 80` |
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