Answer
The abundance of $^{25} Mg$ is $~~9.303\%$
Work Step by Step
Let $a$ be the natural abundance of $^{25} Mg$
Let $b$ be the natural abundance of $^{26} Mg$
We can express $b$ in terms of $a$:
$a+b+0.7899 = 1$
$b = 1-0.7899-a$
$b = 0.2101-a$
We can find the value of $a$:
$(0.7899)(23.985 04~u)+(a)(24.985 84~u)+(b)(25.982 59~u) = 24.312~u$
$(0.7899)(23.985 04)+(a)(24.985 84)+(0.2101-a)(25.982 59) = 24.312$
$(0.7899)(23.985 04)+(a)(24.985 84)+(0.2101)(25.982 59)-25.98259~a = 24.312$
$(0.7899)(23.985 04)+(0.2101)(25.982 59) - 24.312 = 25.98259~a - 24.985 84~a$
$0.092725255 = 0.99675~a$
$a = \frac{0.092725255}{0.99675}$
$a = 0.09303$
The abundance of $^{25} Mg$ is $~~9.303\%$
