In the quantitative estimation of phosphorous by using magnesia mixture, the formula used is:
( w = mass of precipitate formed ; W = mass of organic compound taken)

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Percentage of $P = \frac{31}{222} \times \frac{w \times 100}{W}$

Percentage of $P = \frac{62}{222} \times \frac{W \times 100}{w}$

Percentage of $P = \frac{31}{222} \times \frac{W \times 100}{w}$

Percentage of $P = \frac{62}{222} \times \frac{w \times 100}{W}$

answer is C.

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Detailed Solution

Precipitate is Mg₂P₂O₇.Hence percentage of P = $\frac{2 x GAW of P x weight of ppt .}{GMW of {Mg}_{2} P_{2} O_{7}} x \frac{100}{w e i g h t o f t h e c o m p o u n d}$

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