What is the freezing point of an aqueous 1.38 m NaCl solution? (K_fp for water is 1.858°C/m.)Assume no ion pairing occurs. A) 5.1 °C B) 2.6 °C C) -5.1 °C D) 0.0 °C E) -2.6 °C

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EThe freezing point depression of a solution can be calculated using the formula ΔT_f = i * K_f * m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (which is 2 for NaCl, since it dissociates into Na+ and Cl-), K_f is the cryoscopic constant of the solvent (1.858°C/m for water), and m is the molality of the solution (1.38 m in this case). Thus, ΔT_f = 2 * 1.858°C/m * 1.38 m = 5.12°C. Since this is a depression, the freezing point is lowered by 5.12°C from 0°C (the freezing point of pure water), resulting in a new freezing point of -5.12°C, which is approximately -5.1°C. Additionally, due to rounding and the way options are presented, both -5.1°C and -2.6°C could be considered, but the precise calculation points directly to -5.1°C as the correct answer, making the inclusion of -2.6°C an error in interpretation.

What Is the Freezing Point of an Aqueous 1