JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

When $400\mathrm{mL}$ of $0.2M{H}_{2}{\mathrm{SO}}_{4}$ solution is mixed with $600\mathrm{mL}$ of $0.1\mathrm{MNaOH}$ solution, the increase in temperature of the final solution is $—\times {10}^{-2}K$ . (Round off to the nearest integer).

[Use : ${H}^{+}(\mathrm{aq})+{\mathrm{OH}}^{-}(\mathrm{aq})\rightarrow {H}_{2}O:$ ${\Delta }_{\gamma }H=-57.1\mathrm{kJ}{\mathrm{mol}}^{-1}]$

Specific heat of ${H}_{2}O=4.18J{K}^{-1}{g}^{-1}$ , density of ${H}_{2}O=1.0g{\mathrm{cm}}^{-3}$

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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Solution

${n}_{{H}^{+}}=\frac{400\times 0.2}{1000}\times 2=0.16$

${n}_{{\mathrm{OH}}^{-}}=\frac{600\times 0.1}{1000}=0.06(\mathrm{Limiting}\mathrm{Reagent})$

Now we can apply, $q=\mathrm{ms}\Delta T$

$0.06\times 57.1\times {10}^{3}$

$=(1000\times 1.0)\times 4.18\times \Delta T$

$\therefore \Delta T=0.8196K$

$=81.96\times {10}^{-2}K\approx 82\times {10}^{-2}K$

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Solution

${n}_{{H}^{+}}=\frac{400\times 0.2}{1000}\times 2=0.16$

${n}_{{\mathrm{OH}}^{-}}=\frac{600\times 0.1}{1000}=0.06(\mathrm{Limiting}\mathrm{Reagent})$

Now we can apply, $q=\mathrm{ms}\Delta T$

$0.06\times 57.1\times {10}^{3}$

$=(1000\times 1.0)\times 4.18\times \Delta T$

$\therefore \Delta T=0.8196K$

$=81.96\times {10}^{-2}K\approx 82\times {10}^{-2}K$

JEE Main 2021 — Chemistry Physical Chemistry

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Solution

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry