confusion about titration of ethanoic acid

Hello all, hope someone can help me. Getting really confused with the dilutions and moles.

The investigation is this:
Pipette 10cm3 into a 100cm3 flask and make up to the mark with distilled water. Pipette 25cm3 of the solution into a conical flask and add the indicator. Titrate with 0.1mldm-3 NaOH until a pink colour persists. The titre was 21.1.

The questions are:
1. Calculate the no. of moles of NaOH used:

I can do this, using n = c v its 0.1 x 21.1/1000 = 0.00211

2. Calc the moles of acid in 25cm3 sample of vinegar

Since ratio is 1:1, the moles are the same, so n is 0.00211.

The next questions are where I get confused.

3. Determine the no. of moles of acid in 100cm3 sample of vinegar.

4. Determine the no. of moles of acid in the initial 10cm3 of vinegar

5. Determine the number of moles in 100cm3 of the initial vinegar.

6. Calculate the mass of ethanoic acid in 100cm3 of the initial vinegar

7. Taking 1cm3 of vinegar = 1g, calc the % ethanoic acid in the solution.

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I get really confused at question 3. In my head, we started with 10 cm3 of vinegar, dilluted it to 100 cm3, then took 25 cm3 of this solution.

So if there are 0.0211 moles in 25cm3, I guess in 100cm3 of dilluted soln there are 4 x 0.0211 = 0.00844

4. Determine the no of moles of acid in inital 10 cm3

My notes have this as the same as number 3, 0.00844, but I totally don't understand why, and I get lost here.

Could someone kindly explain (in really simple terms) why it is the same? and the process for the other questions?

Would really appreciate it.

Thanks

They took 25ml out of 100ml so the moles in 100ml = 4 * 0.00211 = 0.00844


The original 10ml was diluted to 100ml so this must be the same as the previous part = 0.00844 mol


I suspect that some info is missing here. Was the 10ml taken from a 100 ml sample originally?