Question

What is the freezing point of a solution prepared by dissolving 50.0g of glycerin in 200g of ethanol?

Answer 1

When a solute is added to a solvent, some properties are affected and these set of properties are called colligative properties. Freezing point depression calculation is used for this type of problem. Calculations are as follows:

ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (50.0/92.09 mol / .200 kg)
ΔT(freezing point)  = 5.04 °C 
Tf - T = 5.04 °C 
T = -5.04 °C

 

Answer 2

Answer:

-120 °C

Explanation:

The freezing point depression is calculated as follows:

ΔT = Kf*m

where ΔT is the difference between the freezing point of the pure solvent and from the solution; kf is a constant (equal to 2 °C/m for ethanol); and m is the molality of the solution.

moles of solute = (50 g)/(92.09 g/mol) = 0.54 mol

molality of the solution = moles of solute / kg of solvent

molality of the solution = 0.54 mol/ 0.2 kg = 2.7 m

ΔT = 2*2.7 = 5.4 °C

The freezing point of ethanol is -114.6 °C, then the freezing point of the solution is -114.6 °C - 5.4 °C =  -120 °C