Question

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 $3 \mathrm{M} \mathrm{HCl}$ that you can prepare using only these two solutions?
(A) $250 \mathrm{ml}$
(B) $500 \mathrm{ml}$
(C) $750 \mathrm{ml}$
(D) $1000 \mathrm{ml}$

Answer 1

Ans : (C)
$\because$ We have to prepare maximum volume solution of $3 \mathrm{~N}$. (3 $\mathrm{N}$ is same as $3 \mathrm{M}$ for $\mathrm{HCl}$ ) Hence we will take al $500 \mathrm{ml}$ of $2 \mathrm{~N}$ solution and $\mathrm{x} \mathrm{ml}$ of $5 \mathrm{~N}$ solution.
$\begin{array}{ll}\therefore \quad & 500 \times 2+x \times 5=3(x+500) \\ & 1000+5 x=3 x+1500 \\ & 2 x=500 \\ & x=250 \mathrm{ml}\end{array}$

Answer 2

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Answer 3

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