Molecular weight titration with NaOH and Fe(II)/Ce(IV) titration

[Radja24]

Hi, I'm taking AP Chem this year, and we just had this big assignment. I got through most of it, but here's some I need a walkthrough with.

Homework Statement

A titration is done using .1302M NaOH to determine the molecular weight of an acid. The acid contains one acidic hydrogen per molecule. If 1.863 grams of the acid requires 70.11mL of the NaOH solution, what is the molecular weight of the acid?

Here's the second one:

Fe (II) + Ce (IV) -----> Fe (III) + Ce (III)

What is the concentration of Fe(II) if it requires 21.35mL of .3136 M Ce (IV) to titrate a 10.00mL aliquot to the endpoint?

Attempt:

1st one, I tried to get moles of NaOH and then multiply it by 110 to get 1 mole, then multiply 1.863 by 110...but that makes no sense.
For the second, I tried using MfVf formula for titration, and before that I tried to get the moles of Ce (Iv) once again, and then find the limiting agent and plug and go from there, but I sincerely am stuck.
These were two ones I couldn't get really get after some nonsense work. Please help...

 

[symbolipoint]

The basic plan for both is answer the question of how many MOLES were titrated?
Each of those analyses use 1:1 mole ratios of titrant to analyte being sought.

(moles per liter of titrant)*(liters of titrant) = moles of titrant

 

[Radja24]

For the second one I did .3136 * .02135 to get .0067 moles Ce (IV)m then multiplied .0067 * .01 Liters to get .000067 moles of titrant, and then I divided that number by .01 L to get .0067 M as my answer