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The natural boron of atomic weight 10.81...

The natural boron of atomic weight 10.81 is found to have two isotopes `B^(10)andB^(11)`. The ratio of abundance of isotopes in natural boron should be

A

`11:10`

B

`81:19`

C

`10:11`

D

`19:81`

Text Solution

Verified by Experts

The correct Answer is:
D
Let abundance of `B^(10)` be `m%`
So, abundance of `B^(11)=(100-m)%`
`therefore 10.81=((10xxm)+11(100-m))/(100)`
or `1081=10m+1100-11m`
or `m=19`
`therefore` Ratio of abundances = `(19)/(100-19)`
`=(19)/(81)`
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