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$19.85 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{NaOH}$ reacts with $20 \mathrm{~mL}$ of $\mathrm{HCl}$ solution for complete neutralisation. The molarity of $\mathrm{HCl}$ solution is
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$0.099$
$\begin{array}{l}
\text { Molarity of base }=\frac{\text { normality }}{\text { Acidity }}=\frac{0.1}{1}=0.1 \\
\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} \\
0.1 \times 19.85=\mathrm{M}_{2} \times 20 \\
\mathrm{M}_{2}=0.09925 \approx 0.099
\end{array}$
\text { Molarity of base }=\frac{\text { normality }}{\text { Acidity }}=\frac{0.1}{1}=0.1 \\
\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} \\
0.1 \times 19.85=\mathrm{M}_{2} \times 20 \\
\mathrm{M}_{2}=0.09925 \approx 0.099
\end{array}$
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