If 30ml of 0.5m koh is needed to neutralize 2m hcl, what was the volume of the acid?

The equation for the reaction between KOH and HCl is as follows
KOH + HCl ---> KCl + H2O
the stoichiometry of KOH to HCl = 1:1
the number of KOH moles reacted = 0.5 mol /1000 cm³ * 30 cm³ 
                                                       = 0.015 mol
the number of HCl moles reacted = number of KOH moles reacted 
therefore HCl moles reacted =  0.015 mol
the molarity of HCl is 0.2 mol/dm³
0.2 mol of HCl in - 1000 cm³
Therefore volume required for 0.015 mol = 1000 cm³ / 0.2 mol * 0.015 mol 
                                                                 = 75 cm³
Therefore 75 cm³ of HCl is required

The correct answer is "atomic weight".

Carbon based or cells?