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.

Temperature of freezing for the solution is. 3.3°C

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

the answer is 303.15 or 303