0.54g of an organic compound containing phosphorous gave 0.888 g of $M g_{2} P_{2} O_{7}$ by the usual analysis. Calculate the percentage of phosphorous in the compound.

85.4%

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45.88%

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27.2%

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12.8%

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Solution

The correct option is C 45.88%
The mass of the compd = 0.54 g
Mass of Mg2P2O7 = 0.88 g
Since 1 mole of $M g_{2} P_{2} O_{7}$ has 2 g atoms of P, or
222 g of $M g_{2} P_{2} O_{7}$ = 62 g of P
So, 0.88 g of $M g_{2} P_{2} O_{7}$ contains P = 62/222 x 0.88 g
So % P present in the compound = $62 / 222 \times 0.88 \times 100 / 0.54$
= 45.88%

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