# What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?

##### 1 Answer

#### Answer:

#### Explanation:

The *short answer* is that the volume of the target solution will be equal to

Think about it like this - you want the concentration of the target solution to be **half** of the concentration of the stock solution.

Since you're keeping the amount of solute **unchanged**, you can say that **doubling the volume** of the solution will result in **half** the initial concentration

Now for the *long answer*.

So, you're dealing with

A *mass by volume* percent concentration is defined as **mass of solute**, usually expressed in *grams*, divided by **volume of solution**, usually expressed in *milliliters*, and multiplied by

#color(blue)("%m/v" = "mass of solute"/"volume of solution" xx 100)#

In your case, the stock solution will contain

#"5.0% g/mL" = m_(KOH)/"125 mL" xx 100#

#m_(KOH) = (5.0"g"/color(red)(cancel(color(black)("mL"))) * 125 color(red)(cancel(color(black)("mL"))))/100 = "6.25 g"#

So, your stock solution contains

This means that you have

#"2.5% g/mL" = "6.25 g"/V_"sol" * 100#

#V_"sol" = (6.25 color(red)(cancel(color(black)("g"))))/(2.5color(red)(cancel(color(black)("g")))/"mL") * 100 = color(green)("250 mL")#