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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 `.^10B` and `.^11B` .The ratio of abundance of isotopes of 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 `x%`.
`therefore" Abundance of "B^(11)=(100-x)%`
`therefore" "10.81=((10xx 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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