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Chemistry · 2021

JEE Main · 27 July 2021, Shift 2 · Q53

When 400 mL of 0.2 M H_2 SO_4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is \ \ \ \ ×…

When $\displaystyle 400$ mL of $\displaystyle 0.2 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is mixed with $\displaystyle 600$ mL of $\displaystyle 0.1$ M NaOH solution, the increase in temperature of the final solution is \_\_\_\_ $\displaystyle \times 10^{-2} \mathrm{~K}$. (Round off to the Nearest Integer). [Use : $\displaystyle \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_2 \mathrm{O}: \Delta_\gamma \mathrm{H}=-57.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Specific heat of $\displaystyle \mathrm{H}_2 \mathrm{O}=4.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1}$ density of $\displaystyle \mathrm{H}_2 \mathrm{O}=1.0 \mathrm{~g} \mathrm{~cm}^{-3}$ Assume no change in volume of solution on mixing.]
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.