# If water is added to 100.0 mL of a 0.150 M sodium hydroxide solution unit the final volume is 150.0 mL, what will the molarity of the diluted solution be?

Mar 15, 2017

The end concentration will be...........$0.100 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$, i.e. $C = \frac{n}{V}$.

So $n = C V$.

Since here $n$ is constant:

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$

And ${C}_{2} = \frac{{C}_{1} {V}_{1}}{V} _ 2$ (and the RHS clearly has the units of concentration, why?)

So ${C}_{2} = \frac{100.0 \cdot \cancel{m L} \times 0.150 \cdot m o l \cdot {L}^{-} 1}{150 \cdot \cancel{m L}} = 0.100 \cdot m o l \cdot {L}^{-} 1$

Here, I did not have to convert the $m L$ volume to litres, because these clearly cancel out in the calculation to give an answer in $m o l \cdot {L}^{-} 1$ as required.........