# How can I calculate the dilution factor using concentration?

##### 1 Answer

Here's how you could do that.

#### Explanation:

You can start by making sure that you understand what it means to **dilute** a solution.

The underlying principle when performing a *dilution* is the fact that the number of moles of solute **must remain constant**.

As you know, the concentration of a solution is defined as the number of moles of solute per **liter** of solution.

#color(blue)("molarity" = "moles of solute"/"liter of solution")#

The purpose of a dilution is to **decrease** the concentration of the solution. Since the number of moles of solute **must remain constant**, the only way to achieve the decrease in concentration is to **increase** the volume of the solution by adding *more solute*.

Now, the dilution factor is simply the ratio between the **final volume** of the diluted solution and the **initial volume** of the concentrated solution.

#color(blue)("D"."F". = V_"final"/V_"initial")#

Let's say that you don't know the values of the two volumes, which we'll call

#c_1 -># the molarity of theconcentratedsolution

#c_2 -># the molarity of thedilutedsolution

Since the number of moles of solute, **constant** for both solutions, you can say that

#c_1 = n/V_1 " "# and#" "c_2 = n/V_2#

Rearrange to get the volumes of the two solutions

#V_1 = n/c_1" "# and#" "V_2 = n/c_2#

This means that the **dilution factor** will be

#"D"."F". = V_2/V_1 = V_2 * 1/V_1 = color(red)(cancel(color(black)(n)))/c_2 * c_1/color(red)(cancel(color(black)(n)))#

#"D"."F". = color(green)(c_1/c_2)#

Since the concentration of the stock solution is **higher** than the concentration of the diluted solution, you can express the former as a **multiple** of the latter

#c_1 = m xx c_2#

The dilution factor will be

#"D"."F". = (m * color(red)(cancel(color(black)(c_2))))/color(red)(cancel(color(black)(c_2))) = m#

You would thus say that *the stock solution was diluted by a factor of*

Alternatively, you can say that you performed a