Find the number of oxygen atoms in one formula unit of $Mg_3(PO_4)_2$.

In $Mg_3(PO_4)_2$:

• There are $2$ phosphate groups, $(PO_4)$
• Each $(PO_4)$ has $4$ oxygen atoms

So total oxygen atoms per formula unit: $2 \times 4 = 8$

Therefore, 1 mole of $Mg_3(PO_4)_2$ contains 8 moles of oxygen atoms.

Let the required moles of $Mg_3(PO_4)_2$ be $x$.

Then moles of oxygen atoms present will be: $8x$

Given that this equals $0.25$ mole: $8x = 0.25$

Solve for $x$: $x = \frac{0.25}{8}$ $x = 0.03125$

In scientific notation: $x = 3.125 \times 10^{-2}$

Match with the options:

• A: $1.25 \times 10^{-2}$
• B: $2.5 \times 10^{-2}$
• C: $0.02$
• D: $3.125 \times 10^{-2}$

So the correct option is: $\boxed{D}$

Comparison with stored correct answer:

Stored correct answer: D

My derived answer is also D.