Verified Solution • 2025–26 Session moderate

QJEEM22Chc08d: JEE Main 2022 Chemistry - Class 11 JEEM22Chc08d Solution: When 600 mL of 0.2 M HNO 3 is mixed with of 0.1 M NaOH solution …

When 600 mL of 0.2 M HNO3 is mixed with 400 mL400 \mathrm{~mL} of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ___________ ×10−2∘C.

(Enthalpy of neutralisation =57 kJ mol−1=57 \mathrm{~kJ} \mathrm{~mol}^{-1} and Specific heat of water =4.2 JK−1 g−1=4.2 \,\mathrm{JK}^{-1} \mathrm{~g}^{-1} ) (Neglect heat capacity of flask)

Step-by-Step Verified Solution

HNO3
600 mL × 0.2 M = 120 m mol

NaOH
400 mL × 0.1 M = 40 m mol

JEE Main 2022 (Online) 29th July Morning Shift Chemistry - Thermodynamics Question 65 English Explanation
Heat liberated from reaction

=40×10−3×57×103 J=40 \times 10^{-3} \times 57 \times 10^{3} \mathrm{~J}

Heat gained by solution =mCΔT=m C \Delta T

m= mass of solution =V×damp;=1000×1amp;=1000 g\begin{aligned}\mathrm{m}=\text { mass of solution }=\mathrm{V} \times \mathrm{d} &=1000 \times 1 \\\\&=1000 \mathrm{~g}\end{aligned}

Heat gained by solution =1000×4.2×ΔT…(2)=1000 \times 4.2 \times \Delta \mathrm{T} \ldots(2)

From (1) and (2)

Heat liberated == Heat gained

40×10−3×57×103=1000×4.2×ΔT40 \times 10^{-3} \times 57 \times 10^{3}=1000 \times 4.2 \times \Delta T

ΔT=54×10−2∘C\Delta T=54 \times 10^{-2}{ }^{\circ} \mathrm{C}

(Rounded off to the nearest integer)

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