Verified Solution • 2025–26 Session moderate

QJEEM21Ch955d: JEE Main 2021 Chemistry - Class 11 JEEM21Ch955d Solution: When 400 mL of 0.2 M H 2 SO 4 solution is mixed with 600 mL of…

When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ ×\times 10 −- 2 K. (Round off to the nearest integer).

[Use : H+ (aq) + OH −- (aq) →\to H2O : Δ\Delta γ\gamma H = −- 57.1 kJ mol −- 1]

Specific heat of H2O = 4.18 J K −- 1 g −- 1

density of H2O = 1.0 g cm −- 3

Assume no change in volume of solution on mixing.

Step-by-Step Verified Solution

nH+=400×0.21000×2=0.16{n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16

nOH−=600×0.11000=0.06{n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06 (L.R.)

Now, heat liberated from reaction = heat gained by solutions

or, 0.06 ×\times 57.1 ×\times 103

= (1000 ×\times 1.0) ×\times 4.18 ×\times Δ\Delta T

∴\therefore Δ\Delta T = 0.8196K

= 81.96 ×\times 10 −- 2 K ≈\approx 82 ×\times 10 −- 2 K

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring