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# How many moles of magnesium phosphate $\text{M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}$, will contain $0.25$moles of oxygen atoms? A. $0.02$B. $3.125\times {{10}^{-2}}$C. $1.25\times {{10}^{-2}}$ D. $2.5\times {{10}^{-2}}$

Last updated date: 13th Jun 2024
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Hint: In order to calculate mole of oxygen we first calculate the number of moles of oxygen in one mole of $\text{M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}$ then we will calculate the moles of $\text{M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}$ in 0.25 mole of oxygen.
Complete step by step solution:
Magnesium phosphate is $\text{M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}$, which means magnesium phosphate has 3 moles of magnesium (Mg), 2 moles of phosphorus (P) and 8 moles of oxygen (O).
So 1 mole of $\text{M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}$, contains & 8 moles of oxygen.
Given $0.25$ moles of oxygen atoms
$\text{1mole M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}\to \text{8 moles oxygen}$
$\text{x moles M}{{\text{g}}_{\text{3}}}{{\text{(P}{{\text{O}}_{\text{4}}}\text{)}}_{\text{2}}}\to 0.25\text{ moles oxygen}$
Number of moles of $\text{M}{{\text{g}}_{3}}{{(P{{O}_{4}})}_{2}}$ having $0.25$ moles of oxygen atoms be x.
Therefore x can be calculated as,
$\text{x=}\dfrac{\text{0}\text{.25 }\!\!\times\!\!\text{ 1moles oxygen}}{\text{8 moles oxygen}}$
$\text{x = 0}\text{.03125}$
That means moles$0.03125$ of magnesium phosphate contains $0.25$ moles of oxygen atoms.

Note: By knowing the moles of any element in 1 mole of the compound, we can find the number of moles of the compound having n moles of the element.