What is the concentration of the base (lioh) in this titration?

a.) 1.20 m
b.) 1.18 m
c.) 30 m
d.) 1.5 m

If the conc. of HCl is 1.5 M, then 1000ml of HCl contain 1.5 mol
                                            let  20ml of HCl (used in titration)contain x
 
                              ⇒ x = (20 ml  × 1.5 mol) ÷1000ml
                                  x = 0.03 mol
   ∴ moles of HCl used to dissolve the LiOH was 0.03 mol

The reaction is characterised by the equation: HCl + LiOH → LiCl  +  H₂O
 
Based on the equation, the mole ratio of HCl : LiOH  is   1  :  1
       ∴ if moles of HCl reacting = 0.03 mol
              then moles of LiOH = 0.03mol

Now, since 25ml of LiOH reacted with the HCl,
then it implies that 25ml of LiOH contain 0.03 mol

        ∴ Let    25ml of LiOH contain 0.03mol 
        and  1000ml of LiOH contain    y
     
                    ⇒ y = (1000 ml  ×  0.03 mol) ÷ 25ml
                        y = 1.2 mol

∴ the concentration of the base (LiOH) in this titration is 1.2 M (OPTION A)