Final answer:
The freezing point depression of a 0.05 m CaCl2 solution would be greater than that of a 0.10 m NaCl solution but less than that of a 0.20 m NaCl solution due to the different number of ions each compound produces upon dissolving in water.
Step-by-step explanation:
You're asking about how the freezing point depression of a 0.05 m CaCl2 solution compares with that of a NaCl solution. The key concept here is understanding colligative properties, which are properties that depend on the number of particles in a solution. Freezing point depression is one such property, and it is affected by the number of dissolved particles in the solution.When NaCl dissolves, it dissociates into two ions: Na+ and Cl−. Conversely, CaCl2 dissociates into three ions: one Ca2+ and two Cl−. So, for every mole of CaCl2 that dissolves, it releases more particles compared to a mole of NaCl.
Given a 0.05 m CaCl2 solution, the effective concentration of particles after dissociation would be 3 times the initial molality, or 0.15 m. This concentration is greater than that for a 0.10 m NaCl solution (which would effectively be 0.20 m after dissociation into two ions per mole) but less than that for a 0.20 m NaCl solution (which would effectively be 0.40 m after dissociation).Therefore, the freezing point depression of a 0.05 m CaCl2 solution would be greater than that for a 0.10 m NaCl solution but less than that for a 0.20 m NaCl solution.