JEE Main 202621 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 21 January
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Practical Chemistry
JEE Main 2026 · 21 January · Evening Shift · Chemistry · Practical Chemistry
By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium...
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 )
Options
- A30
- B50
- C40
- D20
Correct answer
B. 50
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%.
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%.