Question

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

Answer 1

`"20 M"`

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

`V_"diluted"` - the volume of the diluted solution
`V_"concentrated"` - the volume of the 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 `40` times more concentrated than the diluted solution. Since the diluted solution is said to have a molarity of `"0.5 M"`, you can say that the stock solution had 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

`c_1`, `V_1` - the molarity and volume of the concentrated solution
`c_2`, `V_2` - the molarity and volume of the diluted solution

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)|)))`