# 25 mL of a #"10% m/v"# #NH_4Cl# solution is diluted to 250 mL. What is the final concentration of the #NH_4Cl#?

##### 1 Answer

#### Explanation:

You can actually solve this problem without using any formulas, provided that you have a clear understanding of what it means to **dilute** a solution.

The underlying principle behind a *dilution* is that the **number of moles** of solute, i.e. the *mass of solute*, remains **constant**.

The concentration of the original solution **decreases** following a dilution because the **volume** of the solution is increased by adding more *solvent*.

Now, your original solution is **per**

In other words, you get *one part* solute **for every**

Notice that you dilute this solution to a volume that is **ten times** bigger than the initial volume.

Since the mass of solute remained **constant**, you can say that the *diluted* solution contains **one part** solute **for every**

This means that the concentration of the solution **decreased** by a factor of

#"% m/v" = 1/10 * "10%" = color(green)(|bar(ul(color(white)(a/a)1%color(white)(a/a)|)))#

The ratio that exists between the **final volume** and the **initial volume** of the solution is called the **dilution factor**.

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"final"/V_"initial"color(white)(a/a)|)))#

As you can see, the dilution factor allows you to calculate the concentration of the diluted solution by simply looking at the two two volumes.