A dilution lowers a solution's concentration by adding solvent without changing the amount of the solute being diluted.

If the solute amount is unchanged,

$$n_1=n_2.$$

Using $n=cV$ before and after dilution gives

$$\boxed{c_1V_1=c_2V_2}.$$

This relation is often used to prepare a dilute working solution from a more concentrated stock solution.

Example

What volume of a $2.00,\mathrm M$ stock solution is needed to prepare $250.0,\mathrm{mL}$ of a $0.300,\mathrm M$ solution?

Convert consistently or use the same volume unit on both sides:

$$c_1V_1=c_2V_2.$$

Therefore

$$V_1=\frac{(0.300)(250.0,\mathrm{mL})}{2.00}=37.5,\mathrm{mL}.$$

A measured $37.5,\mathrm{mL}$ portion of stock solution is then diluted until the final solution volume is $250.0,\mathrm{mL}$.

This does not mean that exactly $212.5,\mathrm{mL}$ of solvent must always be added: liquid volumes are not guaranteed to be perfectly additive. The preparation is defined by the final volume.

The equation $c_1V_1=c_2V_2$ applies only when the amount of the relevant solute is conserved during the operation. It should not be used unchanged if the solute reacts, precipitates, evaporates or is otherwise gained or lost.