Titration Question

25cm^3 of 0.56M solution of NaOH solution was titrated against a 35gdm^3 of an unknown acid. The titration was repeated several times to give an average volume of 23.85cm^3.
What is the relative formula mass of the acid?
A step by step process on how to answer this question would be much appreciated.

Moles NAOH= (25 X 0.56)/1000
=0.014
Assuming it's a 1:1
Moles acid= 0.014
Need to work out concentration of the acid in moldm-3
Concentration = n x1000/v
= 0.014x1000/23.85
= 0.587 moldm3
Remember that to convert from grams per dm3 to moldm3 you have to divide by the atomic mass.
Here, to work out the Atomic mass you can just divide grams by moles. 35/0.587= 59.6