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. $1.5\,L$ 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)

$0.5L$ of $A+1.5L$ of B

Answer 2

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 $\underset{(For0.5\,N\,HCl)}{\mathop{{{N}_{1}}{{V}_{1}}}}\,+\underset{(For\,0.1\,N\,HCl)}{\mathop{{{N}_{2}}{{V}_{2}}}}\,=\underset{(resulting\,solution)}{\mathop{NV}}\,$ $0.5\times {{V}_{1}}+0.1\times {{V}_{2}}=0.2\times 0.2$ $0.5\,{{V}_{1}}+0.1\,{{V}_{2}}=0.4$ ?(ii) Multiply Eq (i) with 0.5 and then, subtract Eq (ii) from it, $\begin{aligned} & \underline{\begin{aligned} & 0.5{{V}_{1}}+0.5{{V}_{2}}=0.5\times 2\,\,\,\,\,\,\,\,\,...(iii) \\\ & 0.5{{V}_{1}}+0.5{{V}_{2}}=0.4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\\ \end{aligned}} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4{{V}_{2}}=0.6 \\\ \end{aligned}$ ${{V}_{2L}}=1.5L$ ${{V}_{1}}=2-1.5=0.5\,L$ Hence, 0.5 L of A + 1.5 L of B are mixed.