Question

What is the boiling point in ∘C of each of the solutions below? For water, Kb = 0.51(∘C ⋅ kg)/mol.

A solution of 15.0 g of LiCl in 158.0 g of water at 45.0 ∘C, assuming complete dissociation.

Answer 1

Using, Tb - Tb° = i×Kb×m

Where Tb is boiling point of solution

Tb° is boiling point of pure water = 100 °C

Kb = 0.51 °C.Kg/mol

i = 2 = number of ions formed after dissociation

Molality, m = mass of LiCl×1000/molar mass×mass of water

= 15×1000/158×42.394

= 2.239 m

Now, Tb = Tb° + i×Kb×m

= 100 + 2×0.51×2.239

= 100 + 2.284

= 102.28 °C