Question

Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na₂SO₄, and 0.030 mol of MgCl₂, assuming complete dissociation of these electrolytes and ideal solution behavior.

a) 1.33 °C
b) 1.58 °C
c) 1.22 °C
d) 1.95 °C

Answer 1

The boiling point elevation of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation and ideal solution behavior, is approximately 1.78°C.

Step-by-step explanation:

To calculate the boiling point elevation of water containing various solutes, we need to consider the colligative property of boiling point elevation.

First, determine the total number of moles of ions produced by complete dissociation of the solutes. NaCl dissociates into 2 ions, Na2SO4 into 3 ions, and MgCl2 into 3 ions. Total moles of ions = (0.010 mol NaCl × 2) + (0.020 mol Na2SO4 × 3) + (0.030 mol MgCl2 × 3) = 0.20 + 0.060 + 0.090 = 0.35 moles of ions.

The molal concentration (m) is then calculated by dividing total moles of ions by the mass of the solvent in kilograms. Thus m = 0.35 mol / 0.1 kg = 3.5 m.

The boiling point elevation can be calculated using the equation ΔTb = i × Kb × m, where i is the van't Hoff factor (total number of particles after dissociation), Kb is the boiling point elevation constant for water (0.51°C/m), and m is the molality of the solution.

Since the ions fully dissociate and contribute individually, the van't Hoff factor is already included in the calculated molality.

Therefore, the boiling point elevation is ΔTb = 0.51°C/m × 3.5 m = 1.785°C. The boiling point of the solution is approximately 1.78°C higher than the normal boiling point of water.