In 3.10 M KOH solution
            Â
3.10 moles of solute i.e KOH present in 1000 ml solution
            Â
So, 475 ml of solution contain (3.10/1000) × 475
                   Â
Number of moles of KOH i.e 1.4725 number of
                   Â
Moles of KOH
In 12 M KOH solution
             Â
12 moles of solute i.e KOH present in 1000 ml solution
Now, 1.4725 number of moles of KOH
will present in
                                                Â
(1000/12) × 1.4725 ml solution
                                               Â
=122.7 ml solution
Hence, 122.7 ml of 12M KOH solution
needs to be diluted to produce 475 ml of 3.10 KOH solution.