Here's how to calculate the approximate freezing point depression of a 0.020 m aqueous NaBr solution:

Understanding Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the concentration of solute particles in a solution, not their identity. The formula for freezing point depression is:

ΔT_f = K_f * m * i

where:

* ΔT_f is the freezing point depression (change in freezing point)

* K_f is the molal freezing point depression constant for the solvent (water in this case)

* m is the molality of the solution

* i is the van't Hoff factor, representing the number of ions each formula unit of the solute dissociates into in solution.

Calculations

1. K_f for water: The molal freezing point depression constant (K_f) for water is 1.86 °C/m.

2. Molality (m): You've provided the molality of the solution as 0.020 m.

3. Van't Hoff factor (i): NaBr dissociates into two ions in solution (Na⁺ and Br^-), so i = 2.

Putting it all together:

ΔT_f = (1.86 °C/m) * (0.020 m) * (2)

ΔT_f ≈ 0.074 °C

Conclusion

The approximate freezing point depression of a 0.020 m aqueous NaBr solution is about 0.074 °C. This means the freezing point of the solution will be approximately 0.074 °C lower than the freezing point of pure water (0 °C).