Question #0c5a9

1 Answer
Mar 22, 2016


#"3.1% m/v"#


The key to any dilution problem is the fact that the number of moles of solute remains constant.

The solution becomes more diluted, i.e. its concentration decreases, because the volume of the solution increases.

In other words, in order to dilute something, you must make sure that the mass of solute remains constant and that you increase the volume of the solution by adding more solvent.

In your case, the initial solution is said to #"25% m/v"#, which means that you get #"25 g"# of potassium hydroxide for every #"100 mL"# of solution.

Use the given mass by volume percent concentration to determine how many grams of potassium hydroxide you get in that #"15.0-mL"# sample

#15.0 color(red)(cancel(color(black)("mL solution"))) * overbrace("25.0 g KOH"/(100color(red)(cancel(color(black)("mL solution")))))^(color(purple)("25% m/v")) = "3.75 g KOH"#

This is how much solute you have in your initial sample, and how much solute you mmust have in your target solution.

Since mass by volume percent concentration is defined as

#color(blue)(|bar(ul(color(white)(a/a)"% m/v" = "mass of solute"/"100 mL solution" xx 100color(white)(a/a)|)))#

you can say that a #"120.0-mL"# solution that contains #"3.75 g"# of potassium hydroxide will have a mass by volume percent concentration of

#"% m/v" = "3.75 g"/"120 mL" xx 100 = 3.125%#

You should round this off to two sig figs, the number of sig figs you have for the concentration of the initial solution

#"% m/v" = color(green)(|bar(ul(color(white)(a/a)"3.1%"color(white)(a/a)|)))#


Another way to tackle dilution problems is by using the dilution factor, which is the ratio between the final volume of the solution and the initial volume of the sample

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

In your case, the dilution factor is equal to

#"D.F." = (120.0 color(red)(cancel(color(black)("mL"))))/(15.0 color(red)(cancel(color(black)("mL")))) = 8#

This tells you that the initial solution is eight times more concentrated than the target solution.

In other words, the target solution is eight times more diluted than the initial solution.

So, if the target solution is eight times more diluted, that means that its concentration is eight times smaller, so

#"% m/v" = 1/8 * "25%" = color(green)(|bar(ul(color(white)(a/a)"3.1%"color(white)(a/a)|))) -># rounded to two sig figs