2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-40

  • question_answer
    An element X has the following isotopic com- position\[:\]\[{{\,}^{200}}X:90%{{\,}^{199}}X:8.0%{{\,}^{202}}X:2.0%\]. The weighted average atomic mass of the naturally occurring element X is closest to

    A)  201 amu

    B)  202 amu

    C)  199 amu

    D)  200 amu

    Correct Answer: D

    Solution :

     Average isotopic mass of \[X=\frac{200\times 90+199\times 8+202\times 2}{90+8+2}\] \[=\frac{18000+1592+404}{100}=\frac{19996}{100}\] \[=199.96\,\text{amu}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner