JEE Main 2025 · 24 January · morning · Numerical answer · +4 / −1
Xg of benzoic acid on reaction with aq NaHCO₃ released CO₂ that occupied 11.2 L volume at STP.
X is _________ g.
Answer and solution
Answer 61
Solution
Chemistry ma’am · 4 steps
Each COOH group gives exactly one CO₂, so moles of gas tell you moles of acid directly.
- Write the reaction. Benzoic acid is stronger than carbonic acid, so it pushes CO₂ out of bicarbonate. One COOH group releases one CO₂, so the mole ratio is 1:1.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂ + H₂O
- Moles of CO2. At STP one mole of gas occupies 22.4 L. So n(CO₂) = 11.2/22.4 = 0.5 mol, and the acid used is also 0.5 mol.
- Molar mass. Benzoic acid is C₇H₆O₂: 7 × 12 + 6 × 1 + 2 × 16 = 122 g mol⁻¹.
- So. X = 0.5 × 122 = 61 g, so the answer is 61.
The common slip: 60. Using 22.7 L (the molar volume at 1 bar) gives about 60. This question means STP, where one mole is 22.4 L.
The idea to keep. The bicarbonate test releases one CO₂ for every COOH group. Count the acid groups, convert gas volume to moles, and the stoichiometry follows.
Answer from the NTA final answer key. Solution by PrepWiser.