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.

`19:81`

`10 :11`

`15:16`

`81:19`

Text Solution

Verified by Experts

The correct Answer is:

Let % of `B^(10)` atom = x
Average atomic weight `= (10 x+ 11 (100-x))/(100)= 10.81`
`rArr x= 19`
`therefore (""_(5)B^(10))/(""_(5)B^(11))= (19)/(81)`

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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 isotopes of natural boron should be

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In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Relative atomic mass of boron is 10.8 Reason (R) - Boron has two isotopes B-10 and B-11 with percentage abundance of 19.6% and 80.4% respectively.

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