# Question #30f4f

Jun 18, 2017

We WOULD normally use ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, so.........${C}_{2} \cong 0.02 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

The product ${C}_{1} {V}_{1}$ has units $m o l \cdot {L}^{-} 1 \times L$, i.e. you get an answer in $m o l$ as required. Here, we want the $\text{concentration}$, that is given by the quotient $\text{moles"/"volume}$..........

And so here we use this quotient.................

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, ${C}_{2} = \frac{{C}_{1} {V}_{1}}{V} _ 2$,

${C}_{2} = \frac{50.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.195 \cdot m o l \cdot {L}^{-} 1}{5.00 \times {10}^{2} \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= \frac{1}{10} {C}_{1}$

And this makes sense given that we have diluted the volume $\text{tenfold}$.