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An element X has the following istopic ...

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 to

A

`202 am u`

B

`200 am u`

C

`199 am u`

D

`201 am u`

Text Solution

Verified by Experts

The correct Answer is:
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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