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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    Atomic weight of boron is 10.81 and it has two isotopes \[_{5}{{B}^{10}}\] and\[_{5}{{B}^{11}}\]. Then ratio of \[_{\text{5}}{{\text{B}}^{\text{11}}}\text{:}{{\,}_{\text{5}}}{{\text{B}}^{\text{11}}}\] in nature would be :

    A)  81 : 19          

    B)                         15 : 16 

    C)         10 : 11                 

    D)         19 : 81

    Correct Answer: D

    Solution :

    Suppose the percentage of \[{{B}^{10}}\] atoms be Y, then average atomic weight \[=\frac{10y+11\,(100-y)}{100}=10.81\] So,          \[y=19,\]       So,          \[\frac{{{N}_{{{B}^{10}}}}}{{{N}_{{{B}^{11}}}}}=\frac{19}{81}\]


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