# Question #0c5a9

##### 1 Answer

#### Answer:

#### Explanation:

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 **for every**

Use the given *mass by volume percent concentration* to determine how many grams of potassium hydroxide you get in that

#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 m**must 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

#"% 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)|)))#

**ALTERNATIVE APPROACH**

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