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The natural boron is composed of two isotopes `._5B^10` and `._5B^11` having masses `10.003 u` and `11.009 u` resp. Find the relative abundance of each isotope in the natural boron if atomic mass of natural boron is 10.81u. |
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Answer» Let `._(5)B^(10)` be `x%`. Therefore `._(5)B^(11)` will be `(100-x)%`. As average atomic mass = weighted average of the masses of isotopes `:. 10.81 = (10.003x+11.009(100-x))/(100)` Calculate `x = 19.78%` `:. (100 -x) = 100 - 19.78 = 80.22%` |
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