# What volume (in mL's) of a 6.0 M stock solution of HCl should be used to prepare 6.00 L of a 0.250 M HCl solution?

##### 1 Answer

#### Answer:

#### Explanation:

Let's assume for a second that you're not familiar with the equation for dilution calculations.

What would your approach be here?

As you know, *diluting* a solution means keeping the number of moles of *solute* **constant** while **increasing** the total volume of the solution by adding more *solvent*.

Since molarity is defined as moles of solute per liters of solution, diluting a solution will result in a **decrease** in its concentration.

So, if the number of moles of solute **must be constant** in order for a dilution to be performed, you can use the molarity and volume of the *target solution* to figure out how many moles of solute **must be present** in the sample of stock solution you're diluting.

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

#n_(HCl) = "0.250 M" * "6.00 L" = "1.5 moles HCl"#

Now all you have to do is figure out what volume of

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

#V_"stock" = (1.5 color(red)(cancel(color(black)("moles"))))/(6.0 color(red)(cancel(color(black)("moles")))/"L") = "0.25 L"#

Expressed in *milliliters*, the answer will be

#V_"stock" = color(green)("250 mL") -># rounded to two sig figs

This is the principle behind the equation for dilution calculations, which looks like this

#color(blue)(c_1V_1 = c_2V_2)" "# , where

Notice that you actually have

#overbrace(c_1V_1)^(stackrel( color(brown)("moles of solute"))(color(brown)("in stock solution"))) = overbrace(c_2V_2)^(stackrel(color(purple)("moles of solute"))(color(purple)("in target solution")))#

In your case, you'll once again get

#V_1 = c_2/c_1 * V_2#

#V_1 = (0.250 color(red)(cancel(color(black)("M"))))/(6.0color(red)(cancel(color(black)("M")))) * "6.00 L" = "0.25 L" = color(green)("250 L")#