[1800chisaki]

We want to determine the actual molarity of an unknown sample H2SO4. Therefore we titrate the unknown sample with NaOH (1M, f = 1.002).

2 NaOH + 1 H2SO4 → 1 Na2SO4 + 2 H2O

Valid at the equivalence point:

2x #mmol H2SO4 = 1x #mmol NaOH

Suppose you titrate 20.00 ml H2SO4 with 11.00 ml NaOH. What is then the actual molarity of the sulfuric acid (in mmol / ml)?

molarity = n / V

V H2SO4 = 20.00 mL

How do i find n?

n NaOH = m x V

= 1 x 11 = 11 moles

and then 2x #mmol H2SO4 = 1x #mmol NaOH

so 11 x 2 = 22 mmole H2SO4

thus m = 22 / 20 = 1,1 mmoles/mL ?

[chenbeier]

The last step you have to think again.

and then 2x #mmol H2SO4 = 1x #mmol NaOH

How much would be # if x = 1?

22 is wrong.
And also dont forget die titration factors.
The NaOH has 1,002 for instance.