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An element X has the following isotopic ...

An element X has the following isotopic composition :
`.^(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 :

Text Solution

Verified by Experts

The correct Answer is:
200
`barA = sum f_(i)A_(i)` (`f_(i)` = fractional abundance, `A_(i)` = atomic mass)
`=0.90 xx 200 + 0.08 xx 199 + 0.02 xx 202`
`= 180 + 15.92 + 4.04 = 200`
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