# What volume (in L) of a 4.13 M lithium nitrate LiNO_3 solution would be needed to make 195 mL of a 1.05 M solution by dilution?

May 17, 2016

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

${V}_{2} \cong \cdot 50 \cdot m L$

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

Thus $\text{Moles}$ $=$ $\text{Volume "xx" concentration}$

So ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where $C = \text{Concentration; V=Volume}$

And ${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1$ $=$ $\frac{1.05 \cdot m o l \cdot {L}^{-} 1 \times 0.195 \cdot L}{4.13 \cdot m o l \cdot {L}^{-} 1}$ $=$ ??L