Question

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

• A. 5
• B. 20
• C. 10
• D. 15

Answer 1

Answer: (C)

10

Answer 2

The average atomic mass of an element is calculated as the weighted average of the masses of its isotopes.

Given:

• Average atomic mass of Mg = 24.31
• Mole % of $^{24}$Mg = 79%
• Remaining 21 mole % is a mixture of $^{25}$Mg and $^{26}$Mg.

Let the mole % of $^{26}$Mg be $x$. Then, the mole % of $^{25}$Mg will be $(21 - x)$.

The approximate atomic masses of the isotopes are:

• $^{24}$Mg: 24 amu
• $^{25}$Mg: 25 amu
• $^{26}$Mg: 26 amu

Using the formula for average atomic mass:

$$\text{Average Atomic Mass} = \sum (\text{Isotopic Mass} \times \text{Abundance})$$ $$24.31 = \left(24 \times \frac{79}{100}\right) + \left(25 \times \frac{21-x}{100}\right) + \left(26 \times \frac{x}{100}\right)$$

Multiply the entire equation by 100:

$$24.31 \times 100 = (24 \times 79) + (25 \times (21-x)) + (26 \times x)$$ $$2431 = 1896 + (525 - 25x) + 26x$$

Combine the constant terms and the terms with $x$:

$$2431 = (1896 + 525) + (-25x + 26x)$$ $$2431 = 2421 + x$$

Now, solve for $x$:

$$x = 2431 - 2421$$ $$x = 10$$

Therefore, the mole % of $^{26}$Mg is 10%.