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$10 \mathrm{~cm}^{3}$ of $0.1 \mathrm{~N}$ monobasic acid requires $15 \mathrm{~cm}^{3}$ of sodium hydroxide solution whose normality is
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The correct answer is:
$0.066 \mathrm{~N}$
Volume of monobasic acid $=10 \mathrm{~cm}^{3}$
Normality of monobasic acid $=0.1 \mathrm{~N}$
Volume of $\mathrm{NaOH}$ solution $=15 \mathrm{~cm}^{3}$
Normality of $\mathrm{NaOH}$ solution $=$ ?
$$
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (for monobasic acid) } \begin{aligned}
10 \times 0.1 \mathrm{~N} &=15 \times \mathrm{N}_{2} \\
\mathrm{~N}_{2}=\frac{1 \mathrm{~N}}{15} &=0.066 \mathrm{~N}
\end{aligned}
\end{gathered}
$$
Normality of monobasic acid $=0.1 \mathrm{~N}$
Volume of $\mathrm{NaOH}$ solution $=15 \mathrm{~cm}^{3}$
Normality of $\mathrm{NaOH}$ solution $=$ ?
$$
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (for monobasic acid) } \begin{aligned}
10 \times 0.1 \mathrm{~N} &=15 \times \mathrm{N}_{2} \\
\mathrm{~N}_{2}=\frac{1 \mathrm{~N}}{15} &=0.066 \mathrm{~N}
\end{aligned}
\end{gathered}
$$
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