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)

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

Rethink and reinvent