Aconcentrated koh (mw = 56.105 g/mol) solution is 45.0 %wt koh and has a density of 1.45 g/ml. how many milliliters of the concentrated koh solution must be diluted to obtain 250.0 ml of 1.20 m koh?

The wt% of KOH = 45%

This implies that there is 45 g of KOH in 100 g of the solution

Density of the solution is given as 1.45 g/ml

Therefore, the volume corresponding to 100 g of the solution is

= 100 g * 1 ml /1.45 g = 68.97 ml = 0.069 L

Now concentration of the concentrated KOH solution is:

Molarity = moles of KOH/vol of solution

            = (45 g/56.105 g.mol-1)/0.069 L = 11.6 M

Thus,

Initial KOH concentration M1 = 11.6 M

Initial volume = V1

Final concentration M2 = 1.20 M

Final volume V2 = 250 ml

M1*V1= M2*V2

V1 = M2*V2/M1 = 1.20*250/11.6 = 25.9 ml = 26 ml

fe 55.85 g/mol and h2o 18.02 g/mol.

the correct answer to this question is 2