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

Mar 23, 2016

$\text{1% m/v}$

#### 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 $\text{10% m/v}$ ammonium chloride. This essentially means that you get $\text{10 g}$ of ammonium chloride per $\text{100 mL}$ of solution.

In other words, you get one part solute for every $\text{10 mL}$ of solution.

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 $\text{100 mL}$ of solution.

This means that the concentration of the solution decreased by a factor of $10$. As a result, the new concentration will be

"% 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.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F." = V_"final"/V_"initial} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.