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0.62g of an organic compound gave 0.222 Mg₂P₂O₇(Mol.wt = 222amu) by the usual analysis. The percentage of phosphorus in the compound is

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answer is C.

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Detailed Solution

$boxed{% ,,,of,,,P = frac{{62}}{{222}} times frac{{{w_1}}}{w} times 100}$

= $\frac{62}{222} \times \frac{0.222}{0.62} \times 100$ =10

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