Answer:
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
Then, 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 %
