Mole ConceptMedium
Question
By usual analysis, 1.00 g of compound $(X)$ gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound $(X)$ is: (nearest integer)
(Given, molar mass in $g{mol}^{- 1}:O = 16,Mg = 24$, $P = 31$ )
Options
A.50
B.30
C.20
D.40
Solution
% of $P = \frac{n_{{Mg}_{2}P_{2}O_{7}} \times 2 \times 31}{{\text{ }W}_{\text{(unknown compound)~}}} \times 100$
$$\begin{matrix} & \ = \frac{\left( \frac{1.79}{222} \times 2 \times 31 \right)}{1} \times 100 \\ & \ = 49.99\% \approx 50\%. \end{matrix}$$
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