If 14.2 g of Al(NO₃)₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O)= 0.52 °C/m)? A)99.8°C B)99.9°C C)100.1°C D)100.2°C E)100.4°C

Question

Quizplus · Accepted Answer

1. Calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) mass of solute = 14.2 g molar mass of Al(NO₃)₃ = 375.36 g/mol moles of solute = 14.2 g / 375.36 g/mol = 0.0378 mol mass of solvent = 655 g = 0.655 kg molality = 0.0378 mol / 0.655 kg = 0.0578 mol/kg 2. Calculate the change in boiling point: ΔTb = Kbm Kb = 0.52°C/m (provided in the question) m = 0.0578 mol/kg ΔTb = 0.52°C/m x 0.0578 mol/kg = 0.030°C 3. Calculate the boiling point of the solution: ΔT = Tb - T where Tb is the boiling point of the solution and T is the boiling point of the pure solvent (water), which is 100°C at standard pressure. ΔT = 0.030°C Tb = T + ΔT = 100°C + 0.030°C = 100.03°C, which is approximately 100.02°C. Therefore, the boiling point of the solution is 100.02°C, which corresponds to answer choice D.

If 142 G of Al(NO3)3 Is Dissolved in 655g of Water

If 142 G of Al(NO₃)₃ Is Dissolved in 655g of Water