Chemistry · Practical Organic Chemistry

JEE Main 2026 — 21 January, Evening Shift — Question 51

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

Option A:
50
Correct
Option B:
30
Option C:
20
Option D:
40

Answer: A

Step-by-step solution

% of $\mathrm{P}=\frac{\mathrm{n}_{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times 2 \times 31}{\mathrm{~W}_{\text {(unknown compound) }}} \times 100$

\begin{aligned} & =\frac{\left(\frac{1.79}{222} \times 2 \times 31\right)}{1} \times 100 & =49.99 \% \approx 50 \% \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis