Question

What is the freezing point (in K) of a solution made by dissolving 13.136g of CCl₄ in 130.0 g of benzene?  Pure benzene has a freezing point of 5.5°C and a K_f = 5.12 °C/m. Write your answer in Kelvin!

Answer 1

delta Tf = Kf * molality

molality = number of moles of solute / mass of solution in kg

number of moles of solute = 13.136g / 153.82 g/mol = 0.0854 mole

mass of solution in kg = 130.0 g = 0.13 kg

Therefore, molality = 0.0854 mole / 0.13 kg = 0.657 m

delta Tf = 5.12 * 0.657 = 3.36 ⁰C

deltaTf = Tf of solvent - Tf of solution

3.36 = 5.5 - freezing point of solution

freezing point of solution = 5.5 - 3.36 = 2.14 ⁰C = 275.14 K

therefore, freezing point of solution = 275.14 K