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Q.
You are supplied with $500 \,mL$ each of $2N \,HCl$ and $5N \,HCl$. What is the maximum volume of $3M \,HCl$ that you can prepare using only these two solutions?
As per given relation,
$N_{1} \,V_{1}+N_{2} \,V_{2}=N\left(V_{1}+V_{2}\right)$
$N_{1}=2 \,N_{2}=5$
$V_{1}=500 \,V_{2}=$ say $-x$
Thus,
$2 \times 500+5 x=3(x+500)$ or $2 x=500$
i.e. $ x=250$
Hence, maximum volume of
$3\, M\, HCl =500+250=750\, mL$