JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Some Basic Concepts of Chemistry

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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)

(A)50✓ Correct
(B)30
(C)20
(D)40

Answer: (A)

Solution

Phosphorus is estimated as magnesium pyrophosphate,

Mg_{2} P_{2} O_{7}

, of molar mass

2 (24) + 2 (31) + 7 (16) = 222 g mo l^{- 1}

, each formula unit holding two phosphorus atoms. The

1.79 g

precipitate therefore contains

\frac{1.79}{222} \times 2 \times 31 = 0.50 g

of phosphorus. As this came from

1.00 g

of the compound, the percentage of phosphorus is

\frac{0.50}{1.00} \times 100 = 50%

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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