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$25 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{NaOH}$ solution neutralises $12.5 \mathrm{~mL}$ of $\mathrm{HCl}$ solution. The amount of water needed to convert $500 \mathrm{~mL}$ of such $\mathrm{HCl}$ solution to $0.1 \mathrm{~N}$ is
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The correct answer is:
$500 \mathrm{~mL}$e
$\because$ Concentration of $\mathrm{NaOH}\left(N_1\right)=0.1 \mathrm{~N}$
Volume of $\mathrm{NaOH}$ used $\left(V_1\right)=25 \mathrm{~mL}$
Concentration of $\mathrm{HCl}\left(N_2\right)=0.1 \mathrm{~N}$
Volume of $\mathrm{HCl}\left(V_2\right)$ used $=V_2$
and
$N_1 V_1=N_2 V_2$
Thus,
$V_2=\frac{N_1 V_1}{N_2}=\frac{0.1 \times 25}{0.1}$
Volume of $\mathrm{HCl}$ used $\left(V_2\right)=25 \mathrm{~mL}=V_1$
Thus, to convert $500 \mathrm{~mL} \mathrm{HCl}$ solution to $0.1 \mathrm{~N} \mathrm{HCl}$, we need $500 \mathrm{~mL}$ of water.
Volume of $\mathrm{NaOH}$ used $\left(V_1\right)=25 \mathrm{~mL}$
Concentration of $\mathrm{HCl}\left(N_2\right)=0.1 \mathrm{~N}$
Volume of $\mathrm{HCl}\left(V_2\right)$ used $=V_2$
and
$N_1 V_1=N_2 V_2$
Thus,
$V_2=\frac{N_1 V_1}{N_2}=\frac{0.1 \times 25}{0.1}$
Volume of $\mathrm{HCl}$ used $\left(V_2\right)=25 \mathrm{~mL}=V_1$
Thus, to convert $500 \mathrm{~mL} \mathrm{HCl}$ solution to $0.1 \mathrm{~N} \mathrm{HCl}$, we need $500 \mathrm{~mL}$ of water.
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