# how-much-water-would-i-need-to-add-to-500-ml-of-a-2-4-m-kcl-solution-to-make-a-1

##”700 mL”##

Let’s assume that you’re not familiar with the formula for .

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 constant**.

Now, a solution’s 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 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 ##”2.4 M”## to ##”1.0 M”##.

This is exactly how the formula for 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

##c_1##, ##V_1## – the concentration and volume of the initial solution##c_2##, ##V_2## – the concentration and volume of the target solution

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