What is the freezing point of a solution containing 600 g of CHCl₃ and 42.0 g of eucalyptol (C₁₀H₁₈O) , a fragrant substance found in the leaves of eucalyptus trees?
Molar Mass(C₁₀H₁₈O) = 154.25 g/mol, T_f(CHCl₃) = - 63.5 ° C and K_f(CHCl₃) = 4.68 ° C/ m .

Question

Quizplus · Accepted Answer

The freezing point depression of the solution can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (which is 1 for non-electrolytes like eucalyptol), Kf is the cryoscopic constant of the solvent (4.68 °C/m for CHCl3), and m is the molality of the solution. First, calculate the moles of eucalyptol: moles = 42.0 g / 154.25 g/mol = 0.272 moles. Then, calculate the molality: molality = moles of solute / kg of solvent = 0.272 moles / 0.6 kg = 0.453 m. Finally, calculate ΔTf: ΔTf = 1 * 4.68 °C/m * 0.453 m = 2.12 °C. The new freezing point is the original freezing point minus ΔTf: -63.5 °C - 2.12 °C = -65.62 °C, which rounds to -65.6 °C, matching choice A.

What Is the Freezing Point of a Solution Containing 600