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You are supplied with $500 \mathrm{mL}$ each of $2 \mathrm{N} \mathrm{HCl}$ and $5 \mathrm{N} \mathrm{HCl}$. What is the maximum volume of 3M HCl that you can prepare using only these two solutions?
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Verified Answer
The correct answer is:
$750 \mathrm{mL}$
As per given relation,
$\begin{array}{l}
N_{1} V_{1}+N_{2} V_{2}=N\left(V_{1}+V_{2}\right) \\
N_{1}=2 N_{2}=5 \\
V_{1}=500 V_{2}=\operatorname{say}-x
\end{array}$
Thus, $2 \times 500+5 x=3(x+500)$ or $2 x=500^{\circ}$
ie. $x=250$
Hence, maximum volume of
$3 \mathrm{M} \mathrm{HCl}=500+250=750 \mathrm{mL}$
$\begin{array}{l}
N_{1} V_{1}+N_{2} V_{2}=N\left(V_{1}+V_{2}\right) \\
N_{1}=2 N_{2}=5 \\
V_{1}=500 V_{2}=\operatorname{say}-x
\end{array}$
Thus, $2 \times 500+5 x=3(x+500)$ or $2 x=500^{\circ}$
ie. $x=250$
Hence, maximum volume of
$3 \mathrm{M} \mathrm{HCl}=500+250=750 \mathrm{mL}$
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