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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. `19: 81`B. `10 :11`C. `15 :16`D. `81 : 19` |
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Answer» Correct Answer - A Let `._(5)^(10)B` and `._(5)^(11) B` be in the ratio m : n Average atomic weight `10.81 = (mxx 10+ nxx11)/(m +n)rArr (m)/(n)=(0.19)/(0.81) =(19)/(81)`. |
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