JEE Main202127 Jul 2021Evening ShiftChemistryThermodynamics (C)Actual
When 400   mL of 0 . 2   M   H 2 SO 4 solution is mixed with 600   mL of 0 . 1 MNaOH solution, the increase in temperature of the final solution is — × 10 - 2   K . (Round off to the nearest integer). [Use : H + ( aq ) + OH - ( aq ) → H 2 O : Δ γ H = - 57 . 1   kJ   mol - 1 Specific heat of H 2 O = 4 . 18   J   K - 1   g - 1 , density of H 2
Correct answer
0
Step-by-step solution
n H + = 400 × 0 . 2 1000 × 2 = 0 . 16 n OH - = 600 × 0 . 1 1000 = 0 . 06 (   Limiting   Reagent ) Now we can apply, q = ms ∆ T 0 . 06 × 57 . 1 × 10 3 = ( 1000 × 1 . 0 ) × 4 . 18 × ΔT ∴ ΔT = 0 . 8196   K = 81 . 96 × 10 - 2   K ≈ 82 × 10 - 2   K