If you dilute 13.0 ml of the stock solution to a final volume of 0.260 l, what will be the concentration of the diluted solution?

V = 0.260 L = 260 mL

If you give the initial concentration a number, any number, then use C = n/V or n = C*V, one thing you will notice is that n does not change. So your formula becomes

C*V = Cf * Vf

C = 2 mol/L (this number is just made up.)

V = 13 mL

C1 = ? (you have not added anything but water).

V1 = 260 mL

2 * 13 = 260 x

26 = 260 x

x = 0.100 mol/L

the new concentration is 0.1 moL/L The ratio of old to new is 2/0.1 = 20 which means that the old concentration is 20 times the new.

Now we have to do it the more expected way. n is still the same during the dilution.

C = C

V = 13 mL

C1 = ?

V1 = 260 mL

C * 13 = C1 * 260 mL

C / C1 = 260 / 13

C / C1 = 20

The old concentration was 20 times the diluted one. If you are having trouble with this idea, drop me a note. I'll see what will get you to understand it.