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The volume of water to be added to $\frac{N}{2} \mathrm{HCl}$ to prepare $500 \mathrm{~cm}^{3}$ of $\frac{N}{10}$ solution is
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Verified Answer
The correct answer is:
$100 \mathrm{~cm}^{3}$
Moles of $\mathrm{HCl}=\frac{1}{2} \times V \quad\left\{\frac{N}{2} \mathrm{HCl}\right\}$
For preparing $\frac{N}{10}$ solution,
$$
\begin{aligned}
& \frac{1}{10}=\frac{\frac{1}{2} \times V}{500} \\
\therefore \quad V &=\frac{1000}{10}=100 \mathrm{~cm}^{3}
\end{aligned}
$$
For preparing $\frac{N}{10}$ solution,
$$
\begin{aligned}
& \frac{1}{10}=\frac{\frac{1}{2} \times V}{500} \\
\therefore \quad V &=\frac{1000}{10}=100 \mathrm{~cm}^{3}
\end{aligned}
$$
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