7.50gram quantity of a diprotic acid was dissolved in water and made up to exactly 250ml. If 25ml of this solution were neutralized by 11.1 mL of 1.0 M KOH then what is the molar mass of the acid?

Question

I did 7.50/5.55x10^-3

Answer 1

OK EXCEPT that the mass is not 7.5 grams. You titrated only 25 mL aliquot of the original sample of 250 mL; therefore, you titrated only 0.75 g.
Thus 0.75/5.55 x 10^-3 = ?? molar mass

Answer 2

how did you get the 5.55 x 10^-3 ?

Answer 3

.0111/2= .00555