Concentrations: Dilution and mixture

Tutorial 5

    

Progressive exercise Which volumes of HCl $ 12 \; M $ and HCl $ 3 \; M $ must be mixed to give $ 1 \; L $ HCl $ 6 \; M $? Calculation: Indication: Volume of HCl $12\;M = x$: First we calculate .......... the total number of acid moles:
$n_{so}=12\cdot x+3\cdot(1-x)\; mol $ or looking at the mixture: ..........$n_{so}=1\cdot 6=6\; mol $ hence the equation: ..........$12\cdot x+3\cdot(1-x)=6$ with solution: ..........
$x=\frac{1}{3}\; L$ and the answer: You have to mix ..........
$\frac{1}{3} L\, HCl \;12\;M$ with $\frac{2}{3} L\, HCl \;3\;M$