Ure carbon tetrachloride, CCl4, freezes at-23.00 ⁰C and has a kf of 29.8⁰C/m. The latest lot has a freezing point of-23.43⁰C. What is the molality of impurities in thesolvent? If the impurity is largely hexachloroethane, C2Cl6, how many grams of this impurity arethere per kg of CCl4?Once the mass of the impurity is determined, calculate the mass % purity:

a) grams of this impurity per kg of CCl4 = 3.416 g/kg of solvent.

b) mass purity % = 99.66%

Explanation:

Given, the freezing point of pure CCl₄ = - 23°C

Presence of impurities lowers the freezing point to - 23.43°C

The freezing point depression constant, Kբ = 29.8°C/m

The lowered freezing point is related to all the parameters through the relation

ΔT = i Kբ × m

where ΔT is the lowered freezing point, that is, the difference between freezing point of pure substance (T⁰) and freezing point of substance with impurities (T).

i = Van't Hoff factor which measures how much the impurities influence/affect colligative properties (such as freezing point depression) and for most non-electrolytes like this one, it is = 1

Kբ = The freezing point depression constant = 29.8°C/m

m = Molality = ?

T⁰ - T = i Kբ m

- 23 - (-23.43) = 1 × 29.8 × m

m = 0.43/29.8 = 0.0144 mol/kg

Them we're told to calculate impurity of the CCl₄

we convert the Molality to (gram of solute)/(kg of solvent) first

Solute = C₂Cl₆

Molar mass = 236.74 g/mol

So, (molality × molar mass) = (gram of solute)/(kg of solvent)

(gram of solute)/(kg of solvent) = 0.0144 × 236.74 = 3.416 (gram of solute)/(kg of solvent)

Mass purity % = (1000 g of pure substance)/(1000 g of pure substance + mass of impurity in 1000 g of pure substance)

1000 g of solvent contains 3.416 grams of impurities

Mass purity % =100% × 1000/(1003.416)

Mass purity % = 99.66 %