How many moles of magnesium phosphate, $M g_{3} (P O_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

0.02

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3.125 \times 10^{- 2}

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1.25 \times 10^{- 2}

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2.5 \times 10^{- 2}

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Solution

The correct option is B $3.125 \times 10^{- 2}$
1 mole of $M g_{3} (P O_{4})_{2}$
$\Rightarrow$ 3 moles of Mg atom +2 moles of P atom + 8 moles of O atom
8 moles of oxygen atoms are present in = 1 mole of $M g_{3} (P O_{4})_{2}$
Number of moles of $M g_{3} (P O_{4})_{2}$ that contain 0.25 mole of O atoms $= \frac{1 \times 0.25}{8}$
$= 3.125 \times 10^{- 2}$ moles of $M g_{3} (P O_{4})_{2} .$

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