Single Choice

The average atomic mass of a mixture containing 79 mole % of $$_{ }^{24 }{ M }g $$ and remaining 21 mole % of $$_{ }^{ 25 }{ M }g $$ and $$_{ }^{ 26 }{ M }g$$, is 24.31. The % mole of $$^{ 26}{ M }g$$ in the mixture is :

A5
B20
C10
Correct Answer
D15

Solution

Assume mole $$\%$$ of $$ ^{ 26 }{ Mg }$$ is $$x$$.

Then, mole % of $$ ^{ 25 }{ Mg } = 21-x $$
So, avg. atomic mass of mixture $$ = \dfrac { 79\times 24+x\times 26+(21-x)\times 25 }{ 100 } $$ $$=24.31$$ (given)

So, $$x = 10$$ %

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