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An element X has the following isotopic composition: `.^(200)X:90% .^(199)X:8.0% .^(202)X:2.0%` The weight average atomic mass of the naturally occurring element X is closest toA. (a)`201` amuB. (b)202 amuC. (c )199 amuD. (d)200 amu |
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Answer» Correct Answer - d Average isotopic mass of `X=(200xx90+199xx8+202xx2)/(90+8+2)` `=(18000+1592+404)/100=19996/100` `=199.96` amu |
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