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 g mol ⁻¹: O =16, Mg =24, P =31 )

Step-by-step solution

Given: 1.00 g of compound (X) produces 1.79 g of Mg₂P₂O₇. Molar mass of Mg₂P₂O₇ = 2(24) + 2(31) + 7(16) = 222 g/mol. Moles of Mg₂P₂O₇ = 1.79 / 222 = 0.00806 mol. Since each mole of Mg₂P₂O₇ 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%.