By usual analysis, 1.00 g of compound (X ) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X ) is : (nearest integer) (Given, molar mass in gmol^-1 :O=16 , Mg=24 ,P=…

Question

By usual analysis, 1.00 g of compound (X ) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X ) is : (nearest integer) (Given, molar mass in gmol^-1 :O=16 , Mg=24 ,P=31 )

Accepted Answer

Correct answer: B. (b) Given: 1.00 g of compound (X ) produces 1.79 g of Mg_2P_2O_7. Molar mass of Mg_2P_2O_7=2 (24 )+2 (31 )+7 (16 )=222 g/ mol. Moles of Mg_2P_2O_7=1.79/222=0.00806 mol. Since each mole of Mg_2P_2O_7 contains 2 moles of P , moles of P=2 0.00806=0.01612 mol. Mass of P=0.01612 31=0.4997 g ≈0.50 g. Percentage of phosphorus = (0.50/1.00 ) 100=50 \%.

$By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : (nearest integer) (Given, molar mass in gmol^{- 1} : O = 16, Mg = 24, P = 31)$

Solution

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