Question

calculate

a) molality

b) molarity

c) mole fraction

of KI if the density of 20% aqueous KI is 1.202 g/mL?

Answer 1

Molar mass of KI = (1 * 39) + (1 * 127) = 166 g/mol

20 g of KI is present in 100 g of solution, i.e. solution contains 20 g of KI and 80 g of water.

Moles of KI = 20 / 166 = 0.120 moles

(1) Molality = Moles of KI / Mass of water in Kg

= 0.120 / 0.08

= 1.5 molal

(2) Density of solution = 1.202 g/ml

Volume = Mass / Density = 100 / 1.202 g/ml

= 83.19 ml = 83.19 * 10^-3 L

Molarity = Moles of KI / Volume of solution in L

= 0.12 / 83.19 * 10^-3

= 1.45 Molar

(3) Moles of water = 80 / 18 = 4.44 moles

Mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)

= 0.12 / 0.12 + 4.44

= 0.026