Suppose you have an aqueous solution containing 158.2 g KOHper

liter, its density is 1.13 g/cm3. You want to prepare a0.250 molal

solution of KPH, starting with 100.0 mL of the originalsolution.

How much water or solid KOH should be added to the100.0 mL

portion?

1132.8 ml of water

Explanation:

you have an aqueous solution contains 158.2 g KOH per liter

so concentration = 158.2/56 = 2.825M

Molarity = 2.825

that means you have 2.825 moles of KOH in 1.00L solution

Mass of Soluet (KOH) = 152.8g

Volume of solution = 1.00L

density of solution = 1.13g/cm3 = 1.13g/ml

therefore mass of solution = VolumeX density = 1000mL X 1.13g/ml.=1130g

Mass of solvent (water) = mass of solution - mass of solute (KOH) = 1130-152.8 = 997.2g

Molality = moles of solute/mass of solvent (Kg)

=2.825 / (997.2/1000) = 2.832molal

to prepare a 0.250 molal solution of KOH, starting with 100.0ml ofthe orginal solution

0.250*X = 2.832 * 100

X = 1132.8 ml of water you have to add