Respuesta :
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:
-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
