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An element `X` has the following istopic compositon: `.^(200)X(90%), .^(199)X(8.0%), .^(202)X(2.0%)` The weighted average atomic mass of the naturally occuring element `X` is closent toA. `202 am u`B. `200 am u`C. `199 am u`D. `201 am u` |
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Answer» Correct Answer - B Average atomic mass = sum total of (Fractional abundance `xx` Isotopic mass) `= (0.9)(200) + (0.08)(199) + (0.02) (202)` `= 180 + 15.92 + 4.04` `= 199.96 ~~ 200` |
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