[asdfasdfasdf]

Given the following data:

Trial             V_NaOH (mL)
  1                 17.48
  2                 17.24
  3                 17.62

a)  The volume of acetic acid used in each trial was 15.00 mL, and the density of the acetic acid solution is 1.00 g/mL.
b)  Determine which trials should be used in the analysis and which should be left out.  [runs 1 and 3 should be used]
c)  Determine the (average) concentration of the acetic acid solution.  [0.5852 M]
d)  Determine the (average) percent by mass of acetic acid in the solution. [3.51]


The problem is I am unable to find the answer for part C and D mainly.  I understand Part B because runs 1 and 3 are close to each other with a difference of 0.14 while the difference for run 1 and 2 is 0.24 and run 2 and 3 is 0.38; therefore, I assume that run 1 and 3 is the answer.  I assumed that Part A is not really a question; thus, I believed that Part A is some more given info to help solve Part C and D.  Apparently, I am not given the moles of NaOH or anything else.  What you see above is pretty much it.  I have tried doing the question, but I am unable to answer them right; also, I am stumped as I have no idea how to get around this.  I know this is the balanced reaction: NaOH_(aq) + HC₂H₃O_2(aq) -------> NaC₂H₃O_2(aq) + H₂O_(l).  The ones in bold are the answers to Part C and D.  Even with the answers, it is still not of any help to understand the steps to get to the answer.  I have talked a lot, so I guess I'll shut up for now.  Also, I've tried using the search engine but no luck at all.  Thanks in advance to anyone who will bother helping.   

[Borek]

Aren't you given titrant (NaOH) concentration?

[asdfasdfasdf]

Aren't you given titrant (NaOH) concentration?

I honestly don't know.  This is only an exercise I must prepare for my lab.  I think I have to use the concentration of NaOH which I did in my previous question.  The average NaOH concentration from the previous question was 0.5001 M.

[asdfasdfasdf]

-.-  I'm very sorry, but I recently found the solution.  Anyways, thank you for helping out.