Naturally occurring boron consists of two isotopes whose atomic weight are `10.01 and 11.01 `. The atomic weight of the natural boron is `10.81 `. Calculate the percentage of each isotopes in natural boron.
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The correct Answer is:
`20%`
Average atomic weight `=(Sigma" Percentage of an isotope"xx"Atomic weight")/(100)` `rArr 10.81=(10.01x +11.01 (100-x))/(100) implies x=20%` Therefore, natural boron contains `20% (10.01)` isotope and `80%` other isotope.
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