Question

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0.617 g of biphenyl (C12H10) in 25.0 g of cyclohexane? Express the temperature numerically in degrees Celsius.

Answer 1

Temperature of freezing for the solution is. 3.3°C

Step-by-step explanation:

This is an excersise of coligative property (freezing low depression)

ΔT = Kf . m

Mass of solute (biphenyl) → 0.617 g

Mass of solvent (cycloexane) → 25 g

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

Let's find out moles of solute:

Moles = Mass / Molar mass

Moles = 0.617 g / 154 g/m → 0.004 moles

To know molalilty, let's convert 25 g to kg

25 g = 0.025 kg

moles / kg = molality

0.004 m / 0.025 kg = 0.16m

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

6.5°C - T° freezing solution = 20°C/m . 0.16m

6.5°C - T° freezing solution = 3.20°C

6.5°C - 3.20°C = T° freezing solution → °3.3 °C