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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. `15 : 16`B. `19 : 81`C. `81 : 19`D. `20 : 53` |
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Answer» Correct Answer - B `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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