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An element, X has the following isotopic composition:

200X : 90%      199X : 8.0%      202X : 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

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Answer - (d) 200 amu 

 

Solution - 

Average atomic mass = sum of masses of its isotopes multiplied by the percentage/ratio of abundance in nature.

Atomic mass of X = $(200×90)+(199×8)+(202×2)\over100$

                             = $18000+1592+404\over100$

                             = $19996\over100$

Atomic mass of X = 199.96 amu

Approximately = 200 amu

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