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

Atomic weight of boron is `10.81` and it has twoisotopes `._(5)^(10)B` and `._(5)^(11)B`. Then, the ratio of atoms of `._(5)^(10) B` and `._(5)^(11)B` in nature would be

`15 : 16`

`19 : 81`

`81 : 19`

`20 : 53`

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Verified by Experts

The correct Answer is:

`N_(1)/N_(2)=ratio`
Average weight`=(N_(1)W_(1)+N_(2)W_(2))/(N_(1)+N_(2))`
`10.81=(10N_(1)+11N_(2))/(N_(1)+N_(2))`
`10.81N_(1)=10.81N_(2)=10N_(1)+11N_(2)`
`0.81N_(1)=0.19N_(2)implies(N_(1))/(N_(2))=(19)/(81)`

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Atomic weight of boron is `10.81` and it has twoisotopes `._(5)^(10)B` and `._(5)^(11)B`. Then, the ratio of atoms of `._(5)^(10) B` and `._(5)^(11)B` in nature would be

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