Answer:
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
