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The volume of $10 \mathrm{~N}$ and $4 \mathrm{~N} \mathrm{HCl}$ required to make $1 \mathrm{~L}$ of $7 \mathrm{~N} \mathrm{HCl}$ are
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The correct answer is:
$0.50 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.50 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$
Let $V$ litre of $10 \mathrm{~N} \mathrm{HCI}$ be mixed with $(1-V)$ litre of 4 $\mathrm{N} \mathrm{HCl}$ to give $(V+1-V)=1 \mathrm{~L}$ of $7 \mathrm{~N} \mathrm{HCl}$.
$$
\begin{aligned}
N_{1} V_{1}+N_{2} V_{2} &=N V \\
10 V+4(1-V) &=7 \times 1 \\
10 V+4-4 V &=7 \\
6 V &=7-4 \\
V &=\frac{3}{6}=0.50 \mathrm{~L}
\end{aligned}
$$
Volume of $10 \mathrm{~N} \mathrm{HCI}=0.50 \mathrm{~L}$
Volume of $4 \mathrm{~N} \mathrm{HCl}=1-0.50=0.50 \mathrm{~L}$
$$
\begin{aligned}
N_{1} V_{1}+N_{2} V_{2} &=N V \\
10 V+4(1-V) &=7 \times 1 \\
10 V+4-4 V &=7 \\
6 V &=7-4 \\
V &=\frac{3}{6}=0.50 \mathrm{~L}
\end{aligned}
$$
Volume of $10 \mathrm{~N} \mathrm{HCI}=0.50 \mathrm{~L}$
Volume of $4 \mathrm{~N} \mathrm{HCl}=1-0.50=0.50 \mathrm{~L}$
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