# How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

##### 1 Answer

#### Answer:

#### Explanation:

Let's assume that you're *not familiar* with the formula for **dilution calculations**.

The underlying concept of a **dilution** is that you can **decrease** the concentration of a solution by **increasing its volume** while **keeping the number of moles of solute constant**.

Now, a solution's **molarity** tells you the number of moles of solute present in *one liter* of that solution.

#color(blue)(c = n_"solute"/V_"solution")#

In a dilution, you know that the number of moles of solute **must remain constant**. This means that you can use the molarity and volume of the initial solution to figure out how many moles it contains

#color(blue)(c = n/V implies n = c xx V)#

#n_(KCl) = 2.4"moles"/color(red)(cancel(color(black)("L"))) * 500 * 10^(-3)color(red)(cancel(color(black)("L"))) = "1.2 moles KCl"#

This is exactly how many moles of potassium chloride must be present in the target solution, which means that you have

#color(blue)(c = n/V implies V = n/c)#

#V_"target" = (1.2color(red)(cancel(color(black)("moles"))))/(1.0color(red)(cancel(color(black)("moles")))/"L") = "1.2 L"#

So, you need the volume of the target solution to be equal to

#V_"target" = "1.2 L" = 1.2 * 10^3"mL" = "1200 mL"#

This means that you must add

#V_"added" = "1200 mL" - "500 mL" = color(green)("700 mL")#

to your initial solution to get its molarity down from

This is exactly how the formula for dilution calculations works

#color(blue)(overbrace(c_1 xx V_1)^(color(brown)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(brown)("moles of solute in diluted solution")))#

Here