In the estimation of phosphorus, 0.35 g of an organic compound was heated with fuming HNO3, the phosphoric acid formed was precipitated as MgNH4PO4 and ignited to Mg2P2O7, which weighed 0.2212 g. Using Mg = 24, P = 31, O = 16, the percentage of phosphorus is

JEE Advanced levelSingle correct optionDifficulty 3 of 3si190

In the estimation of phosphorus, 0.35 g of an organic compound was heated with fuming $HNO X_{3}$ , the phosphoric acid formed was precipitated as $MgNH X_{4} PO X_{4}$ and ignited to $Mg X_{2} P X_{2} O X_{7}$ , which weighed 0.2212 g. Using Mg = 24, P = 31, O = 16, the percentage of phosphorus is

Mg X_{2} P X_{2} O X_{7}

has molar mass 48 + 62 + 112 = 222 and contains 2 P (62 g). Mass of P =

\frac{62}{222} \times 0.2212 = 0.06178

g. %P =

\frac{0.06178}{0.35} \times 100 = 17.65%