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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 :

A

201 amu

B

202 amu

C

199 amu

D

200 amu

Text Solution

Verified by Experts

The correct Answer is:
D
Average atomic weight
`=("% of I"xxA_(1)+"% of II"xxA_(2)+"% of III"xxA_(3))/(100)`
`=(90xx200+8xx199+2xx202)/(100)=199.96=200`
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