What was the molar mass of the unkown acid

a 3.857 g sample of an unkown monoprotic acid was dissolved and diluted to 100.0 mL. A 25.0 mL portion of the acid solution was titrated using 23.61 mL of 0.20 M KOH. What was the molar mass of the unkown acid?

A monoprotic acid has only one H+, so the balanced equation for the reaction is

HX + KOH ---> KX + H2O
1 mole KOH neutralises 1 mole of the acid

moles KOH used = molarity x litres
= 0.20 M x 0.02361 L
= 0.004722 mol KOH

1 mole KOH reacts with 1 mole acid, therefore mole acid = moles KOH
= 0.004722 mol

So there are 0.004722 moles acid in 25.0 ml of the acid solution.

You dissolved the 3.857 g of solution in 100 ml, but only titrated 25.0 ml of this. Thus you only titrated 1/4 of the total moles of acid.

Thus actual moles acid = 4 x 0.004722 mol
= 0.018888 mol

3.857 g acid contains 0.018888 moles

molar mass = mass / moles
= 3.857 g / 0.018888 mol
= 204 g/mol