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The volume in $\mathrm{mL}$ of $0.1 \mathrm{M}$ solution of $\mathrm{NaOH}$ required to completely neutralise $100 \mathrm{~mL}$ of $0.3 \mathrm{M}$ solution of $\mathrm{H}_3 \mathrm{PO}_3$ is
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The correct answer is:
600
$\because$ Phosphorus acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$ is a dibasic acid.
$\begin{aligned}
0.3 \mathrm{M} \mathrm{H}_3 \mathrm{PO}_3 & =0.6 \mathrm{~N} \mathrm{H}_3 \mathrm{PO}_3 \\
N_1 V_1 & =N_2 V_2 \\
0.1 \times V_1 & =0.6 \times 100 \\
V_1 & =600 \mathrm{~mL}
\end{aligned}
$
$\begin{aligned}
0.3 \mathrm{M} \mathrm{H}_3 \mathrm{PO}_3 & =0.6 \mathrm{~N} \mathrm{H}_3 \mathrm{PO}_3 \\
N_1 V_1 & =N_2 V_2 \\
0.1 \times V_1 & =0.6 \times 100 \\
V_1 & =600 \mathrm{~mL}
\end{aligned}
$
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