What is the freezing point of a 1.8 m solution of CBr₄ in C₆H₆? (K_f for C₆H₆ = 4.90°C/m; Freezing point of C₆H₆ = 5.51°C)

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The freezing point depression of a solution can be calculated using the formula ΔT_f = i * K_f * m, where ΔT_f is the freezing point depression, i is the van't Hoff factor (number of particles the solute dissociates into), K_f is the cryoscopic constant, and m is the molality of the solution. For CBr4, i = 1 (it does not dissociate in solution), K_f for C6H6 (benzene) is 4.90°C/m, and the molality is given as 1.8 m. Thus, ΔT_f = 1 * 4.90°C/m * 1.8 m = 8.82°C. The freezing point of the solution is then the freezing point of pure benzene minus ΔT_f: 5.51°C - 8.82°C = -3.31°C.

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