# What is the molarity of a stock solution if 10 mL is diluted to 400 mL with a concentration of 0.5M?

##### 1 Answer

#### Answer:

#### Explanation:

You can solve this problem by calculating the **dilution factor** associated with your dilution.

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"concentrated"color(white)(a/a)|)))#

Here

**diluted solution**

**concentrated solution**

In essence, the *dilution factor* will tell you how **concentrated** the stock solution was compared with the dilute solution.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"stock" = "D.F." xx c_"diluted")color(white)(a/a)|)))#

In your case, you'll have

#"D.F." = (400 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 40#

This means that the stock solution is **times more concentrated** than the diluted solution. Since the diluted solution is said to have a molarity of

#c_"stock" = 40 * "0.5 M" = color(green)(|bar(ul(color(white)(a/a)"20 M"color(white)(a/a)|)))#

The dilution factor method works because it's based on the *underlying principle* of a dilution, i.e. the concentration of the solution is **decreased** by **Increasing** the volume of the solution while keeping the *number of moles* of solute **constant**.

The equation for dilution calculations looks like this

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

Here

This equation can be rearranged as

#c_1/c_2 = V_2/V_1#

This is where the dilution factor comes into play

#"D.F." = V_2/V_1#

Plug this into the above equation to get

#"D.F." = c_1/c_2 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(c_1 = "D.F." xx c_2)color(white)(a/a)|)))#