A .682 g sample of an unknown weak monoprotic acid, HA, was dissolved in sufficient water to make 50mL of solution and was titrated with a .135 M NaOH solution. After the addition of 10.6mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 27.4mL of the .135M NaOH.

1. Calc. the # of moles of acid in the original solution.
2. Calc. the molar mass of the acid.
3. Calc. molarity of the unreacted HA remaining in solution at pH of 5.65.
4. Calc. the [H3O+] at pH of 5.65.
5. Calc the value of the dissociation constant Ka of the acid.
6. Find the pK range of an indicator that would be appropriate for this titration.

I don't want to duplicate what you already know. Tell us what you don't understand about the problem and we can help you through it.

How do I determine the # of moles if I don't know the formula of the acid?

The whole concept of titration is that you use a volume of a standard base (when titrating an acid) and the indicator turns color when you are at the equilvalence point. At that point the mols of the standard base equals the mols of the unknown acid ( this is a monoprotic acid--stated in the problem).
So mols NaOH = M x L = xx
mols unknown acid = the same number, xx.
M NaOH = 0.135
L NaOH to reach eq. pt. = 0.0274.
Post your work if you get stuck.

Okay, I got .003699 mol acid, do I just put the .682 grams over the moles I got?

Your answer is correct for part a.
You also are correct the molar mass = g/mols. The problem forgot to say, and should have said, that the acid was a pure sample. However, yes, that will get part b.

I know that molarity is mol per liter but how do I find the mol of the unreacted acid?

To find the moles of unreacted acid, you need to use the concept of stoichiometry. Since the acid is monoprotic, it can only donate one proton (H+). From the titration, you know that 10.6 mL of the NaOH solution was required to reach a pH of 5.65, which means that at that point, all the acid had not reacted.

To find the moles of unreacted acid, you can calculate the moles of NaOH that has reacted. Since the acid and base react in a 1:1 ratio, the moles of NaOH is equivalent to the moles of unreacted acid.

Moles of NaOH reacted = Molarity of NaOH x Volume of NaOH (in liters)

Moles of unreacted acid = Moles of NaOH reacted

You already know the molarity of NaOH is 0.135 M, and the volume of NaOH required to reach a pH of 5.65 is 10.6 mL. To convert mL to liters, divide by 1000.

Moles of NaOH reacted = 0.135 M x 10.6 mL / 1000 mL

Once you have calculated the moles of NaOH reacted, you have the moles of unreacted acid.