Question

What is the freezing point (in K) of a solution made by dissolving 19.831g 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 with 4 sig figs!

Answer 1

Use depression in freezing point equation;

dTf = i × Kf × m

Where , dTf = depression in freezing point of solvent after adding solute ( freezing point of pure solvent - freezing point of solution)

i = Van't Hoff factor of solute (no. Of ions into which solute dissociates) , i=1 for CCl4(as non-electrolyte) .

Kf = molal depression in freezing point constant of solvent .

m = molality of solution (moles of solute/mass of solvent in kg)

Here for molality, moles of solute = mass/molar mass = 19.831g/153.82g/mol = 0.129 moles CCl4 .

molality (m) = (0.129moles/0.130kg) = 0.992 m

Now , dTf = 1 × 5.12°C/m × 0.992m = 5.079 °C

Thus freezing point of solution = freezing point of pure benzene - dTf = 5.5°C - 5.079°C = 0.421°C = (273.15k+0.421) = 273.57Kelvin = 273.6 K (rounded to 4 sig figs. ) .