nH+=1000400×0.2×2=0.16
nOH−=1000600×0.1=0.06(LimitingReagent)
Now we can apply, q=msΔT
0.06×57.1×103
=(1000×1.0)×4.18×ΔT
∴ΔT=0.8196K
=81.96×10−2K≈82×10−2K
When 400 mL of 0.2MH_2 SO_4 solution is mixed with 600 mL of 0.1 MNaOH solution, the increase in temperature of the final solution is —× 10^-2K.…
JEE Main 2021 — Chemistry Physical Chemistry
2021integermedium
When 400mL of 0.2MH2SO4 solution is mixed with 600mL of 0.1MNaOH solution, the increase in temperature of the final solution is —×10−2K. (Round off to the nearest integer).
[Use :H+(aq)+OH−(aq)→H2O:ΔγH=−57.1kJmol−1]
Specific heat of H2O=4.18JK−1g−1 , density of H2O=1.0gcm−3
Assume no change in volume of solution on mixing.
Official previous-year question
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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