When 400 mL of 0.2 M H SO solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is _____ × 10 K. (Round off to the nearest integer). Specific heat of

Question

When 400 mL of 0.2 M H SO solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is _____ × 10 K. (Round off to the nearest integer). Specific heat of H O = 4.18 J K g Density of H O = 1.0 g cm Assume no change in volume of solution on mixing. 2 4 –2 2 –1 –1 2 –3

Prashant Kotian · Accepted Answer

Correct answer: 82.00. **Step 1: Determine the limiting reagent and moles of reactants** H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O Moles of H₂SO₄ = 0.2 M × 0.4 L = 0.08 mol Moles of NaOH = 0.1 M × 0.6 L = 0.06 mol NaOH is the limiting reagent (requires 0.16 mol H₂SO₄ but only 0.06 mol available). **Step 2: Calculate moles of water formed** From stoichiometry: 2 mol NaOH produces 2 mol H₂O Moles of H₂O formed = 0.06 mol **Step 3: Calculate total mass of solution** Total volume = 400 + 600 = 1000 mL Mass of solution = 1000 mL × 1.0 g/mL = 1000 g **Step 4: Use the enthalpy of neutralization** For strong acid-strong base neutralization: ΔH = −57.3 kJ/mol of H₂O formed Heat released = 0.06 mol × 57,300 J/mol = 3,438 J **Step 5: Calculate temperature change** q = mcΔT 3,438 = 1000 × 4.18 × ΔT ΔT = 3,438/4,180 = 0.823 K **Step 6: Express in scientific notation** ΔT = 0.823 K = 82.3 × 10⁻² K ≈ 82 × 10⁻² K Therefore, the answer is 82.00.

💡 Solution & Explanation

Practice 22,000+ questions like this