# What volume OF SOLVENT must we add to a 500*mL solution of 1.2*mol*L^-1 concentration to dilute the concentration to 1.0*mol*L^-1?

Mar 2, 2017

Look at this dimensionally...........we must add $100 \cdot m L$ to the initial volume, to make up to a volume of $600 \cdot m L$...........

#### Explanation:

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, if we multiply the DIMENSIONS of concentration and volume, i.e.:

$\text{Concentration"xx"Volume} \equiv m o l \cdot \cancel{{L}^{-} 1} \times \cancel{L} = m o l$, i.e. $C V \equiv \text{moles, i.e. amount of substance. }$

And so, we solve for ${V}_{2} = \frac{{C}_{1} {V}_{1}}{C} _ 2$ (and CLEARLY this has the units of volume, as we require.)

And so..........,

${V}_{2} = \frac{1.2 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1} \times 500 \cdot \cancel{m L} \times {10}^{-} 3 \cdot L \cdot \cancel{m {L}^{-} 1}}{1.0 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}}$

$= 0.60 \cdot L$

And so we ADD $100 \cdot m L$ to the original solution.........

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