How many moles of magnesium phosphate, ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$ will contain 0.25 moles of oxygen atoms?

8 oxygen atoms are present in molecule of $\mathrm{Mg}{\left({\mathrm{PO}}_4\right)}_2 = 1$

$0.25\times {\mathrm{NA}}_{\mathrm{A}}$ oxygen atom are present in molecule of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2=\frac{1}{8}\times 0.25{\mathrm{N}}_{\mathrm{A}}$

So, molecules of  ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$ required $=\frac{0.25}{8}{\mathrm{N}}_{\mathrm{A}}$

$\mathrm{Moles}\mathrm{of}{\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2=\frac{\frac{0.25}{8}{\mathrm{N}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{A}}}=\frac{0.25}{8}{\mathrm{N}}_{\mathrm{A}}\times \frac{1}{{\mathrm{N}}_{\mathrm{A}}}=0.03125=3.125\times {10}^{-2}$