Question

A student enters the lab and determines the freezing point of pure liquid to be 35.8 ºC. A nonelectrolyte unknown substance is added to the liquid, and the freezing point of the solution is determined to be 33.2 ºC. If the freezing point depression constant for the solvent is 3.60 ºC/molal, what is the molality of the solution? (Do not enter units with your answer. Express your answer to the correct number of significant figures.)

Answer 1

The molality is 0.722 molal

Step-by-step explanation:

Step 1: Data given

The freezing point of pure liquid = 35.8 ºC

The freezing point of the solution is determined to be 33.2 ºC

the freezing point depression constant for the solvent is 3.60 ºC/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ with ΔT = the freezing point depression = 35.8-33.2 = 2.6 °C

⇒ with i = the van't Hoff factor = 1

⇒ with Kf = the freezing point depression constant = 3.60 °C/m

⇒ with m= the molality

2.60 °C = 1* 3.60 °C/m * m

molality = 2.60 /3.60

Molality = 0.722 molal

The molality is 0.722 molal

Answer 2

The molality of the solutionis 0.72

Step-by-step explanation:

Freezing point depression → ΔT = Kf . m

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

ΔT = 35.8°C - 33.2°C → 2.6°C

Let's replace the data given → 2.6 = 3.60 . m

m = 0.72