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Question

# In a gravimetric determination of P, an aqueous solution of NaH2PO4 is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate, Mg2P2O7 which is weighed. A solution of NaH2PO4 yielded 0.222 g of Mg2P2O7. What weight of NaH2PO4 is present originally? (Atomic mass: Na=23 u, Mg=24 u, P=31 u, N=14 u, O=16 u, H=1u)

A
0.24 g
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B
0.12g
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C
0.18 g
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D
0.36 g
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Solution

## The correct option is A 0.24 g P atom remains conserved in the reaction according to POAC. Molar mass of Mg2P2O7 =[24×2+2×31+7×16]=222 g/mol Mass of Mg2P2O7=0.222 g Moles of Mg2P2O7=0.222222=0.001 mol Moles of phosphorous =2×0.001=0.002 mol =Moles of NaH2PO4 Mass of NaH2PO4=0.002×[23+2+31+4×16] =0.002×120=0.24 g

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