See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Step 1: Identify Gas A. Benzoic acid (Ph–COOH) reacts with NaHCO3 (a mild base). Carboxylic acids are acidic enough to react with sodium bicarbonate, releasing CO2 gas. Reaction: Ph–COOH + NaHCO3 → Ph–COONa + H2O + CO2↑ Gas A = CO2, molecular mass = 44 g/mol. Step 2: Identify Gas B. Phenol (Ph–OH) reacts with sodium metal (Na). Phenol reacts with Na to release H2 gas. Reaction: 2 Ph–OH + 2 Na → 2 Ph–ONa + H2↑ Gas B = H2, molecular mass = 2 g/mol. Step 3: Note on the superscript '14' above NaHCO3. This appears to be a question number or label (question 14 in a series), not a modifier that changes the chemistry. The reaction still produces CO2. Step 4: Calculate the sum. Molecular mass of A (CO2) = 44 Molecular mass of B (H2) = 2 A + B = 44 + 2 = 46 However, the given answer is 48. Re-examining: if the superscript '14' indicates that 14C (carbon-14) is used in the carboxyl group of benzoic acid, then CO2 formed would contain 14C instead of 12C. The molecular mass of 14CO2 = 14 + 16 + 16 = 46. That still gives 46 + 2 = 48. With carbon-14 in the CO2: mass of 14CO2 = 14 + 32 = 46. Sum = 46 + 2 = 48. So Gas A = 14CO2 (molecular mass 46), Gas B = H2 (molecular mass 2), and their sum = 48. Therefore, the correct answer is 48.