Final answer:
To find the boiling point elevation for a 0.90% NaCl solution with a freezing point depression of 0.52°C, one would calculate the molality of the solution and then apply the boiling point elevation equation using the constant for water and the vant Hoff factor for NaCl. The calculated boiling point elevation, added to the water's normal boiling point, yields the solution's boiling point of approximately 100.143°C.
Step-by-step explanation:
To calculate the boiling point elevation (Liso) of a 0.90% w/v (0.154 M) NaCl solution with a given freezing point depression of 0.52°C, one can use the colligative property equations. First, we can determine the molal concentration (molality) of the solution by using the provided molarity and weight/volume percentage. Once the molality is established, we apply the freezing point depression constant for water (Kf = 1.86 °C kg/mol) along with the vant Hoff factor for NaCl (i = 2) since NaCl dissociates into two ions, Na+ and Cl-. The colligative properties formula for freezing point depression is ΔTf = i · Kf · m. Solving for molality (m) using the given freezing point depression (ΔTf = 0.52°C) would give us m = (0.52 °C) / (1.86 °C kg/mol · 2), which is approximately 0.140 mol/kg.
Assuming that the boiling point elevation and freezing point depression have the same magnitudes but opposite signs, we can use this molality to calculate the boiling point elevation using the boiling point elevation constant for water (Kb = 0.512 °C kg/mol). Therefore, the boiling point elevation would be ΔTb = i · Kb · m, which in this case is: ΔTb = 2 · 0.512 °C kg/mol · 0.140 mol/kg = 0.143 °C. Adding this value to the normal boiling point of water (100°C), we get a final boiling point of approximately 100.143°C.