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
Mar 23, 2016

#"1% m/v"#


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 #"10% m/v"# ammonium chloride. This essentially means that you get #"10 g"# of ammonium chloride per #"100 mL"# of solution.

In other words, you get one part solute for every #"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 #"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.

#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.