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NEET Sample Paper NEET Sample Test Paper-81

  • question_answer
    Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. The percentage of each isotope in natural boron is

    A)  20, 80             

    B)  40, 60

    C)  60, 40             

    D)  80, 20

    Correct Answer: A

    Solution :

    Let \[%\] of isotope with atomic weight \[10.01 = x%\]of isotope with atomic weight \[11.01 = \left( 100 - x \right)\] Since, atomic weight \[=\,\,\frac{x\,\,\times \,\,10.01+\left( 100-x \right)\,\,\times \,\,11.01}{100}\] \[10.81\,\,=\,\,\frac{x \times \,\,10.01+\left( 100-x \right)\,\,\times \,\,11.01}{100}\] \[x=20\] Hence, \[%\] of isotope with atomic weight \[10.01\text{ }=\text{ }20\] % of isotope with atomic weight 11.01 \[=100-20\] \[=\,\,80\]


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