HX + NaOH -----> NaX + H2O
The reaction occurs on a 1 mole:1 mole basis, so the number of moles of acid consumed (na) is equal to the number of moles of base consumed (nb). If concentration is given the symbol c, and volume the symbol V, then n = cV. Accordingly …
na = nb
… so ...
caVa = cbVb
From your data, and my assumption about the base concentration:
ca = ?
Va = 25.0 mL
cb = 0.0500 mol/L
Vb = 7.50 mL
Therefore …
(25.0 mL)(ca) = (7.50 mL)(0.0500 mol/L)
ca = (7.50 mL)(0.0500 mol/L) / (25.0 mL)
= 0.0150 mol/L
You originally had 0.100 g of the acid dissolved to form 100 mL of solution. So the concentration of the acid was 1.00 g/L. But we just calculated that the concentration is 0.0150 mol/L. So 1.00 g of the acid is 0.0150 mol.
Molar mass of acid = mass / no, of moles
= 1.00 g / 0.0150 mol
= 66.6 g/mol
So the molar mass of the acid is 66.6 g/mol.
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That’s the procedure. I stress that I just dreamt up the NaOH concentration of 0.0500 mol/L : as far as I know, there is no solid monoprotic acid with a molar mass of 66.6 g/mol, or anywhere near it. But, if you substitute the actual concentration of the NaOH that you used, you’ll come up with an answer.
If you’re dealing with a diprotic or triprotic acid, you can no longer say that na = nb: but I won’t go into that hypothetical situation.
I hope that this is of some assistance to you and, if nothing else, persuades you that the information you supplied is insufficient to solve the problem. God luck!