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Atomic weight of boron is 10.81 and it h...

Atomic weight of boron is `10.81` and it has two isotopes `._5 B^10` and `._5 B^11`. Then ratio of `._5 B^10` in nature would be.

A

`19 : 81`

B

`10 : 11`

C

`15 : 16`

D

`81 : 19`

Text Solution

Verified by Experts

The correct Answer is:
A
Let the percentage of `B^(10)` atoms be x , then average atomic weight
`= (10 x + 11 (100 - x))/(100) = 10.81 implies x = 19 therefore (N_(B) 10)/(N_(B) 11) = (19)/(81)`
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