What volume of 0.123 M NaOH, in mililiters, contains 25.0g of NaOH

I'm kinda lost on how to solve this..
I did..25g/40g to get the moles of NaOH which I got .625mol of NaOH..then I did .123 mol/liters (molarity) multiplied by .625 ( mols of NaOH) & got .076875 liters..then I did a conversion to turn liters into milliliters getting 77 millliters ( when reduced to 2 sig figs)

good first step
25g / 40g/mole = 0.625moles NaOH

second step is incorrect
M = moles / volume
0.123M = 0.625moles / vNaOH
vNaOH = 0.625moles / 0.123M = 5.081L or 5081ml

you multiplied moles/L x moles, this gave you the value of 0.076L but the units would be moles^2/L
i find that if i write down the units while doing the calculations so that i can determine the units of the answer, i find where i went wrong when the wrong units come out.

Atomic weights: Na=23, O=16, H=1, NaOH=40

Let the solution be called S.

25.0gNaOH x 1molNaOH/40gOH x 1000mLS/0.123molNaOH = 5080 mL solution to three significant figures

Everything falls right out when you use the factor label method