JEE Main 2026 — Practical Chemistry Question with Solution
JEE Main 2026 (21 January Shift 2)
Question
By usual analysis, 1.00 g of compound gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound is : (nearest integer) (Given, molar mass in )
Choose an option
Show full solutionCorrect option: B
Correct answer
B50
Step-by-step explanation
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%.
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%.
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This is a previous-year question from JEE Main 2026, covering the Practical Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.