Question

A solution is prepared by dissolving 7.00 g of glycerin (C3H8O3) in 201 g of ethanol (C2H5OH). The freezing point of the solution is ________°C. The freezing point of pure ethanol is -114.6°C at 1 atm. The molal-freezing-point-depression constant (Kf) for ethanol is 1.99°C/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.

Answer 1

Answer:

The freezing point of the solution is -115.35 °C

Explanation:

Let's apply the coligative property of freezing point depression to solve this.

In this case solute is non-electrolytic so, we don't apply Van't Hoff factor.

ΔT = Kf. m

ΔT = T° freezing of solvent pure - T° freezing of solution.

Kf = molal-freezing-point-depression constant

m = molality ( moles of solute in 1kg of solvent)

Let's calculate molality

In 201 g of solvent we have 7 g of solute

In 1000 g we have __ ( 1000 .7) / 201 g = 34.8 g

Now let's convert to moles

Mass / Molar mass → 34.8 g / 92.09 g/m = 0.377 moles

ΔT = 1.99 °C/m . 0.377 m

ΔT = 0.75°C

ΔT = T° freezing of solvent pure - T° freezing of solution.

0.75°C = -114.6 °C - x

x = -114.6°C - 0.75°C

x = -115.35°C