# What volume of water must be added to 57*mL of 0.500*mol*L^-1 solution to make 285*mL of 0.100*mol*L^-1 solution?

Feb 13, 2017

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, and thus we must add $228 \cdot m L$ to the mother solution.

#### Explanation:

The product ${C}_{1} {V}_{1}$ gives the starting molar quantity; this is clear from the dimensional product, i.e. $m o l \cdot {L}^{-} 1 \times L = m o l$ as req'd.

From the given relationship,

${V}_{2} = \frac{{C}_{1} {V}_{1}}{C} _ 2 = \frac{57.0 \times {10}^{-} 3 L \times 0.500 \cdot \cancel{m o l \cdot {L}^{-} 1}}{0.100 \cdot \cancel{m o l \cdot {L}^{-} 1}} = 285 \cdot m L$

But we already have a $57 \cdot m L$ volume, so we add appropriately.