The average atomic mass of a mixture containing 79 mole percent of $^{24} Mg$ and remaining 21 mole percent of $^{25} Mg$ and $^{26} Mg$ , is 24.31. Mole percent of $^{26} Mg$ is :

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Solution

The correct option is C 10
Average atomic mass = $24.31$ %.
The mole percentage of $^{24} Mg$ is $79$ %.
Let the mole percentage of $^{26} Mg$ be $x$ .
The mole percentage of $^{25} Mg$ will be $21 - x$ .
$24.31 = \frac{0.79 \times 24 + 0.01 x \times 26 + 0.01 \times (21 - x) \times 25}{0.79 + 0.01 x + 0.01 (21 - x)}$ .
Hence, $x = 10$ .

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The average atomic mass of a mixture containing 79 mole percent of 24Mg and remaining 21 mole percent of 25Mg and 26Mg, is 24.31. Mole percent of 26Mg is :

The correct option is C 10Average atomic mass = 24.31 %.The mole percentage of 24Mg is 79%.Let the mole percentage of 26Mg be x.The mole percentage of 25Mg will be 21−x.24.31=0.79×24+0.01x×26+0.01×(21−x)×250.79+0.01x+0.01(21−x).Hence, x=10.

The average atomic mass of a mixture containing 79 mole percent of $^{24} Mg$ and remaining 21 mole percent of $^{25} Mg$ and $^{26} Mg$ , is 24.31. Mole percent of $^{26} Mg$ is :