Haha thanks for all your help and different ways to approaching the question.
I have another question:
A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?
The equation for the neutralisation reaction from the info is this:
HNO3 + KOH ---> KNO3 +H2O
The mole ratio between HNO3 and KOH is 1:1 meaning that basically 1 mol of HNO3 will react with 1 mol of KOH. Keep this ratio in mind.
Transform everything to moles for calculation:
n(KOH) in moles = 0.60 * 0.050 = 0.030mol KOH reacted
Remember the mole ratio? We now know that:
n(HNO3) = n(KOH) = 0.030mol due to the 1:1 ratio
Now this 0.030mol is from 0.025L of solution. Hence in 0.100 L of solution we use a multiplication factor of 4 to get the amount in mol in the 100mL.
n(HNO3) in 100.0mL = 4 * 0.030mol = 0.120 mol
This amount is the total mol of HNO3 from the very start (which was originally in only 10mL of solution). Therefore we now can easily calculate the concentration of the original HNO3.
C(HNO3) original = 0.120 / 0.010 = 12M concentration for the nitric acid.
(sheesh that's some strong nitric acid...now I am less sure my answer is right)