What is the freezing point of a solution made with 1.17 mol of CHCl₃ in 530.0...

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Question

What is the freezing point of a solution made with 1.17 mol of CHCl₃ in 530.0 g of CCl₄ (Kf = 29.8 °C/m, Tf = -22.9 °C)?

Answer 1

Answer #1

moles CHCl₃ = 1.17 mol

mass CCl₄ = 530.0 g

mass CCl₄ = 530.0 g * (1 kg / 1000)

mass CCl₄ = 0.5300 kg

molality CHCl₃ = (mass moles CHCl₃) / (mass CCl₄ in kg)

molality CHCl₃ = (1.17 mol) / (0.5300 kg)

molality CHCl₃ = 2.21 m

Decrease in freezing point of solvent = (K_f) * (molality of solute)

Decrease in freezing point of CCl₄ = (K_f) * (molality CHCl₃)

Decrease in freezing point of CCl₄ = (29.8 ^oC/m) * (2.21 m)

Decrease in freezing point of CCl₄ = 65.8 ^oC

Freezing point of solution = (freezing point of pure solvent) - (Decrease in freezing point)

Freezing point of solution = (-22.9 ^oC) - (65.8 ^oC)

Freezing point of solution = -88.7 ^oC

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Answer 2

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