Calculate the freezing point of a solution made from 22.0 g of octane (C₈H₁₈) dissolved in 148.0 g of benzene. Benzene freezes at 5.50^$\circ$C and its K_f value is 5.12^$\circ$C/m. A) -1.16^$\circ$C B) 0.98^$\circ$C C) 5.49^$\circ$C D) 6.66^$\circ$C E) 12.2^$\circ$C

Question

Quizplus · Accepted Answer

First, we need to calculate the freezing point depression (ΔTf) caused by the solution:
ΔTf = Kf * molality
molality = moles of solute / mass of solvent (kg)
mol. mass of octane (C8H18) = 114.23 g/mol
moles of octane = 22.0 g / 114.23 g/mol = 0.1925 mol
mass of solvent (benzene) = 148.0 g / 1000 = 0.1480 kg
molality = 0.1925 mol / 0.1480 kg = 1.2992 mol/kg

ΔTf = 5.12°C/m * 1.2992 mol/kg = 6.65°C
(Note: we convert Celsius to Fahrenheit in the end)

The freezing point of the solution is:
Freezing point = Freezing point of pure solvent - ΔTf
Freezing point = 5.50°F - (6.65°C * 1.8°F/°C) ≈ -1.16°F

Therefore, the best choice is A, which matches the calculated freezing point.

Calculate the Freezing Point of a Solution Made from 22 ∘\circ∘

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