Chemistry · 2022
JEE Main · 28 July 2022, Shift 2 · Q51
2 L of 0.2 M H_2 SO_4 is reacted with 2 L of 0.1 M NaOH solution, the molarity of the resulting product Na_2 SO_4 in the solution is \ \ \ \…
$\displaystyle 2$ L of $\displaystyle 0.2 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ is reacted with $\displaystyle 2$ L of $\displaystyle 0.1$ M NaOH solution, the molarity of the resulting product $\displaystyle \mathrm{Na}_2 \mathrm{SO}_4$ in the solution is \_\_\_\_ millimolar. (Nearest integer)
Official answer
25
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.