Amberlite Resin

More precisely, if g 0.1500 of an unknown sample was passed through a column of Amberlite IR-120 (+) resin and titrated with NaOH 9.75 M. If 0.2050 ml of titrant is required to phenolphthalein reach criterion, what is the molecular weight of the unknown substance?

I will make the assumption that the resin has created something with H + 1, which reacted with Na0H

IR-120 - H + Na0H ===> IR + H0H-120-Na

Now all we need to know is the number of moles of Na0H.

C = 0.2050
n =???
LV = 9.75 mL = 9.75/1000 = 0.00975
n = 0.2050 * 0.00975 = 0.001999 mol.

0.1500 grams of the chemical produced that number of moles.

n = 0.001999 moles
since mass = 0.1500 g
MM =??
MM = given mass / moles
= MM = 0.1500/0.001999 75.0469 grams / mole

I will give it to someone else about it.