Verified Solution • 2025–26 Session moderate

QJEEM21Chc31c: JEE Main 2021 Chemistry - Class 11 JEEM21Chc31c Solution: 200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The…

200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is −- 57.1 kJ. The increase in temperature in ∘^\circ C of the system on mixing is x ×\times 10 −- 2. The value of x is ___________. (Nearest integer)

[Given : Specific heat of water = 4.18 J g −- 1 K −- 1, Density of water = 1.00 g cm −- 3]

[Assume no volume change on mixing)

Step-by-Step Verified Solution

⇒\Rightarrow Millimoles of HCl = 200 ×\times 0.2 = 40

⇒\Rightarrow Millimoles of NaOH = 300 ×\times 0.1 = 30

⇒\Rightarrow Heat released = (301000×57.1×1000)\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right) = 1713 J

⇒\Rightarrow Mass of solution = 500 ml ×\times 1 gm/ml = 500 gm

⇒\Rightarrow ΔT=qm×c=1713J500g×4.18Jg−K\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}} = 0.8196 K

= 81.96 ×\times 10 −- 2 K

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