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$2.5 \mathrm{~cm}^{3}$ of $0.2 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution is diluted to $0.5$ $\mathrm{dm}^{3}$. Find normality of the diluted solution.
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Verified Answer
The correct answer is:
$0.002 \mathrm{~N}$
Given, volume of $\mathrm{H}_{2} \mathrm{SO}_{4}$ before dilution, $V_{1}=2.5 \mathrm{~cm}^{3}$
$$
=2.5 \times 10^{-3} \mathrm{dm}^{3}
$$
volume of $\mathrm{II}_{2} \mathrm{SO}_{4}$ after dilution, $V_{2}-0.5 \mathrm{dm}^{3}$
Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ before dilution $=0.2 \mathrm{M}$
$$
=0.4 \mathrm{~N} \quad[\because 1 \mathrm{~N}=2 \mathrm{M}]
$$
Concentration (normality) of $\mathrm{H}_{2} \mathrm{SO}_{4}$ after dilution, $N_{2}=$ ? Normality equation is
$$
N_{1} V_{1} \quad=\quad N_{2} V_{2}
$$
(before dilution) $\quad$ (after dilution)
$$
\begin{aligned}
2.5 \times 10^{-3} \times 0.4 &=N_{2} \times 0.5 \\
N_{2} &=\frac{2.5 \times 10^{-3} \times 0.4}{0.5} \\
&=2 \times 10^{-3}=0.002 \mathrm{~N}
\end{aligned}
$$
$$
=2.5 \times 10^{-3} \mathrm{dm}^{3}
$$
volume of $\mathrm{II}_{2} \mathrm{SO}_{4}$ after dilution, $V_{2}-0.5 \mathrm{dm}^{3}$
Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ before dilution $=0.2 \mathrm{M}$
$$
=0.4 \mathrm{~N} \quad[\because 1 \mathrm{~N}=2 \mathrm{M}]
$$
Concentration (normality) of $\mathrm{H}_{2} \mathrm{SO}_{4}$ after dilution, $N_{2}=$ ? Normality equation is
$$
N_{1} V_{1} \quad=\quad N_{2} V_{2}
$$
(before dilution) $\quad$ (after dilution)
$$
\begin{aligned}
2.5 \times 10^{-3} \times 0.4 &=N_{2} \times 0.5 \\
N_{2} &=\frac{2.5 \times 10^{-3} \times 0.4}{0.5} \\
&=2 \times 10^{-3}=0.002 \mathrm{~N}
\end{aligned}
$$
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