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The natural boron of atomic weight 10.81 is found to have two isotopes .^10B and .^11B .The ratio of abundance of isotopesof natural boron should be |
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Answer» Solution :Let ABUNDANCE of `B^10` be x% `therefore` Abundance of `B^11` = (100-x)% `therefore 10.81=((10 xx x)+11(100-x))/100` or 1081=10x + 1100 - 11x or x=19 `therefore` Ratio of abundance =`19/(100-19)=19/81` |
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