1.

An element X has the following isotopic composition, .^(200)X:90% .^(199)X:8% .^(2002)X:2% the weighted average atomic mass of the naturally-occurring element 'X' is closed to:

Answer»

201amu
202amu
199amu
200amu

Solution :AVERAGE ATOMIC MASS of
`X=[(99)/(100)xx200]+[(8)/(100)xx199]+[(2)/(100)xx202]`
`=199.96` amu =200amu.


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