Question

A solution is made by dissolving 3.8 moles of sodium chloride (NaCl) in 185 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

Answer 1

The boiling point of the solution can be calculated by the equation of the colligative property named boilng point elevation.

The formula for the boiling point elevation is:

ΔTb = i * Kb * m

Where:

i = van't Hoof factor
Kb = molal boiling point constant
m = molality.

Solution:

1) i is 2 because the NaCl is ionice and dissociates into two ions.

2) Kb is given = 0.51 °C/m

3) m = moles of solute / kg of solvent

moles of solute is given = 3.8

kg of solvent = 185 g * 1 kg/1000g = 0.185 kg

m = 3.8 moles / 0.185 kg = 20.54 m

4) ΔTb = 2 * 0.51°C/m * 20.54m = 20.9 °C

5) The boiling point is the normal boiling point of water (100°C) plus the increase:

Tb = 100°C + 20.9°C = 120.9°C

Answer: 120.9°C.