milimole 040HCl+030NaOH⟶30−NaCl+−−H2O ΔHneut =−57.1KJ
ΔH=[−57.1×30×10−3×103]J=1713J
q=m⋅s.ΔT
1713=500×4.18×ΔT
ΔT=0.8196K=81.96×10−2K≈82×10−2K
200 mL of 0.2M HCl is mixed with 300 mL of 0.1M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in…
JEE Main 2021 — Chemistry Physical Chemistry
2021integermedium
200mL of 0.2MHCl is mixed with 300mL of 0.1MNaOH. The molar heat of neutralization of this reaction is −57.1kJ. The increase in temperature in C∘ of the system on mixing is x×10−2. The value of x is (Nearest integer)
[Given: Specific heat of water =4.18Jg−1K−1
Density of water=1.00gcm−3]
(Assume no volume change on mixing)
Official previous-year question
Held on 27 Aug 2021 · Verified 6 Jul 2026.
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