Determine the freezing point of a solution of 24.0 grams of carbon tetrachloride, CCl4 (n=1.00), dissolved in 1500 grams of benzene (Kf=5.12) which has a freezing point of 5.48 °C

Answer 1

Freezing point depression refers to the lowering of the freezing point of solvents upon the addition of solutes.

It is a colligative property of solutions that is generally proportional to the molality of the added solute. The depression in the freezing point of a solution can be described by the following formula:.

ΔT_f = i×K_f×m

Where

1. ΔT_f is the freezing point depression. =Tf-Tf* = (T-5.48) °C = (Tf and T*f are the freezing points of the solution and pure solvent, respectively).
2. 
   
3. i is the Van't Hoff factor. For non-ionic solutes, i.e., nonelectrolytes, i = 1, as there is only one "dissociated" particle per dissolved particle.
4. K_f is the cryoscopic constant, and Kf=5.12^∘C⋅kg/mol is the freezing point depression constant for benzene, the solvent.
5. m is the molality.

According to Formula, first we need to calculate the molality of CCl4 solution.

Wt of carbon tetrachloride (CCl4) (solute) = 24.0 grams

M.Wt. of carbon tetrachloride (CCl4) = 153.82 g / mol where n = w/mw = 24g g/153.82 g/mol mol= 0.156 moles

wt of benzene (solvent) = 1500g =1.5kg

Molality (m) = n (moles) / w (kg) = 0.156 moles / 1.5kg = 0.104 moles/kg

Therefore, the molality of CCl4 solution is 0.104 moles/kg.

T-5.48 = 1X5.12 ^∘C kg/mol X0.104 mol/kg

(T-5.48)^∘C = 0.532 ^∘C = this is the amount of freezing point depression.

T = (0.532+5.48)^∘C = 6.012^∘C

Therefore, the freezing point of the solution is Tf = 6.012 °C.