The normality of $\mathrm{H}_2 \mathrm{SO}_4$ in the solution obtained on mixing 100 mL of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ with 50 mL of 0.1 M NaOH is $\_\_\_\_$ $\times 10^{-1} \mathrm{~N}$. (Nearest Integer)
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Solution
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4=100 \times 0.1 \times 2=20$
No. of equivalents of $\mathrm{NaOH}=50 \times 0.1=5$
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4$ left $=20-5=15$
$$
\begin{aligned}
\Rightarrow & 150 \times x=15 \\
& x=\frac{1}{10}=0.1 \mathrm{~N}=1 \times 10^{-1} \mathrm{~N}
\end{aligned}
$$
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