Titrations - working out the concentration

hi im stuck on this titration

When a solution of HCL is titrated against 0.9M NaOH, 23cm^3 of the HCL was needed to neutralise 25cm3 of the alkali.

NaOH + HCL -> NaCl + H20

i've woked out the number of moles of alkali used is 2.25 X 10 ^-2 moles,
and the number of moles of acid used is 2.25 X 10 ^-2 moles, and the concentration of the acid in mol dm^-3 to be 0.978 mol dm^-3, but i dont know how to work out the concentration of the acid in g dm^-3

please help

multiplying by the RMM of HCL worked

one more question

Calculate the molarity of the first named solution
25cm^3 of H(subscript x)A react with 25cm^3 of 0.2mol dm^-3 to give Na(subscript 2)A

HxA + 2NaOH ---> Na2A + 2H2O
Moles of NaOH = Conc * Volume = 0.2 * 25/1000 = 0.005
Ratio of NaOH to HxA is 2:1
Moles of HxA = 0.005/2 = 0.0025
Conc of HxA = Moles/Volume = 0.0025/(25/1000) = 0.1M