A 1.5 mol L-1 potassium hydroxide solution has a density of 1.33 g mL -1. What is the molality of the solution?

Answer 1

Molality = moles solute / kg solvent

So, we need to know moles of KOH present, and we also need to know kg of water present.

We begin by assuming we have 1000 ml (1 liter) of solution, and by finding molar mass of KOH = 56.1 g/mol)

1.5 M = 1.5 mol KOH / L

1.5 mol KOH/L x 56.1 g/mol = 84.15 g KOH present

To find mass (kg) water, we subtract mass of KOH from total mass:

total mass of solution = 1000 ml x 1.33 g/ml = 1330 g

mass of water = 1330 g - 84.15 g = 1246 g H2O = 1.246 kg

We now have moles of KOH = 1.5 moles

We also now have kg H2O = 1.246 kg

Solve for molality: 1.5 moles / 1.246 kg = 1.2 m