1.

The natural boron of atomic weight 10.81 is found to have two isotopes .^10B and .^11B .The ratio of abundance of isotopesof natural boron should be

Answer»

`11:10`
`81:19`
`10:11`
`19:81`

Solution :Let ABUNDANCE of `B^10` be x%
`therefore` Abundance of `B^11` = (100-x)%
`therefore 10.81=((10 xx x)+11(100-x))/100`
or 1081=10x + 1100 - 11x or x=19
`therefore` Ratio of abundance =`19/(100-19)=19/81`


Discussion

No Comment Found

Related InterviewSolutions