Question

What volume of water must I evaporate from 10.00 L of a 0.135 mol/kg solution of potassium chloride to get a 0.450 mol/kg solution?

Answer 1

You would have to evaporate 2.21 L of water.

This question involves molality. You can read about molality at

http://socratic.org/questions/what-is-molality

1. Calculate the volume of water in the 0.350 mol/kg solution.

We need to know the density of the 0.350 mol/kg solution. I will assume that the density is 1.017 g/mL.

Mass of solution = 10 000 mL soln × `(1.017"g soln")/(1"mL soln")` = 10 170 g soln

0.350 mol KCl × `(74.55"g KCl")/(1"mol KCl")` = 26.1 g KCl

∴ A 0.350 mol/kg solution of KCl contains 26.1 g KCl in 1000 g H₂O or 26.1 g KCl in 1026.1 g solution.

Mass of KCl = 10 170 g solution × `(26.1"g KCl")/(1026.1"g soln")` = 259 g KCl

Mass of water = 10170 g soln – 259 g KCl = 9910 g water

If the temperature of the water is 20 °C, its density is 0.9982 g/mL.

Volume of water = 9910 g × `(1"mL")/(0.9982"g")` = 9930 mL

2. Calculate the mass of water in the 0.450 mol/kg solution.

You still have 259 g of KCl in the solution.

259 g KCl × `(1"mol KCl")/(74.55"g KCl")` = 3.47 mol KCl

To make a 0.450 mol/kg solution, you will need

3.47 mol KCl × `(1000"g H₂O")/(0.450"mol KCl")` = 7710 g water

Volume of water = 7710 g H₂O × `(1"mL H₂O")/(0.9982"g H₂O")` = 7720 mL

3. Calculate the volume of water to evaporate

Volume = 9930 mL – 7720 mL = 2210 mL = 2.21 L

You will have to evaporate 2.21 L of water.

If your instructor gave different densities for the solutions, you will have to recalculate.