Question

The freezing point of ethanol (C2H5OH) is -114.6 °C. The molal freezing point depression constant for ethanol is 2.00 °C/m. What is the freezing point (°C) of a solution prepared by dissolving 50.0 g of glycerin (C3H8O3, a nonelectrolyte) in 200 g of ethanol

Answer 1

Answer:

Freezing point of the solution = -120.02  °C

Explanation:

When a non-volatile solute is added in a volatile solvent, then vapour pressure of the pure solvent will decrease which results in depression in freezing point.

Molal freezing point depression constant for ethanol = 2.0 °C/m

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal freezing point depression constant

m = Molality

$m = \frac{Mass of solute in g}{Molecular mass of solute\times Mass of solvent in Kg}$

Mass of glycerin (Solute) = 50.0 g

Molecular mass of glycerin = 92.09

Mass of ethanol (Solvent) = 200 g = 0.200 Kg

$m = \frac{50.0}{92.09 \times 0.200}$

m = 2.71

Now, put the value of molality,  molal freezing point depression constant in the formula,

$\Delta T_f = K_f \times m$

$\Delta T_f = 2.00 \times 2.71$= 5.42  °C

Freezing point of solvent - freezing point of solution = ΔTf

-114.6 °C - freezing point of solution = 5.42  °C

freezing point of solution = -5.42 - 114.6

                                           = -120.02  °C

Answer 2

The Freezing point of the solution is -120.02  °C

Calculation of freezing point:

But first molarity should be determined

Since

Mass of glycerin (Solute) = 50.0 g

The molecular mass of glycerin = 92.09

Mass of ethanol (Solvent) = 200 g = 0.200 Kg

So, molarity should be

$= 50 \div (92.09 \times 0.200)$

= 2.71

Now freezing point should be

$= (2.00 \times 2.71) - 114.60$

= 5.42 - 114.60

= -120.02°C

learn more about freezing point here: https://brainly.com/question/21526256