Question

Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL

Answer 1

Basis: 1 L of the substance.
(1.202 g/mL) x (1000 mL) = 1202 g
mass solute = (1202 g) x 0.2 = 240.2 g
mass solvent = 1202 g x 0.8 = 961.6 g
moles KI = (240.2 g) x (1 mole / 166 g) = 1.45 moles
moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent
= 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
= 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
= 1.45 moles / (1.45 moles + 53.42 moles) = 0.0264