2 L of $0.2 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ is reacted with 2 L of 0.1 M NaOH solution, the molarity of the resulting product $\mathrm{Na}_2 \mathrm{SO}_4$ in the solution is $\_\_\_\_$ millimolar. (Nearest integer).
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