0.12g of an organic compound gave 0.22 ${Mg}_{2} P_{2} O_{7}$ by the usual analysis. The percentage of phosphorus in the compound is

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38.75

60.92

51.20

15.23

answer is C.

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Detailed Solution

It is given to find out the percentage of phosphorus in the compound if

0.12g of an organic compound gave 0.22 g of magnesium pyrophosphate.

Explanation
1 mol i.e. 222g of Mg2P2O7 contains phosphorous content = 62 g
Therefore, 1 g of Mg2P2O7 contains phosphorous content = $\frac{62}{222}$ g
Therefore, 0.22 g Mg2P2O7 contains phosphorous content = $\frac{62}{222} \times 0.22$ g
Thus, this is the amount of phosphorous present in 0.12 g of the organic compound.
Therefore, the percentage of phosphorous in the compound is = $\frac{62}{222} \times 0.22 \times \frac{1}{0.12} \times 100 = 51.16 % = 51.20 % (approx)$