Question

The volumes of two $\text{HCl}$ solutions A(0.5 N) and B(0.1 N) to be mixed for preparing 2 L of 0.2 N HCl solution are

• A. $0.5L$ of $A+1.5L$ of B
• B. $1.5L$ of $A+0.5L$ of B
• C. $1L$ of $A+1L$ of B
• D. $0.75L$ of $A+1.25L$ of B

Answer 1

Answer: (A)

Let the volume of 0.5 N HCl solution $\text{=}{\text{V}}_{\text{1}}$ and the volume of 0.1 N HCl solution $\text{=}{\text{V}}_2$ Total volume of the mixture = 2 L ${\text{V}}_1+V_2=2L$ ?(i) We know that $\underset{(For0.5NHCl)}{N_1{V}_1}+\underset{(For0.1NHCl)}{N_2{V}_2}=\underset{(resultingsolution)}{\mathrm{NV}}$ $0.5\times V_1+0.1\times V_2=0.2\times 0.2$ $0.5V_1+0.1V_2=0.4$ ?(ii) Multiply Eq (i) with 0.5 and then, subtract Eq (ii) from it, $\begin{array}{rl}& \underline{\begin{array}{rl}& 0.5V_1+0.5V_2=0.5\times 2...(iii)\\ & 0.5V_1+0.5V_2=0.4...(ii)\end{array}}\\ & 0.4V_2=0.6\end{array}$ $V_{2L}=1.5L$ $V_1=2-1.5=0.5L$ Hence, 0.5 L of A + 1.5 L of B are mixed.