Question

What is the molality of a water solution that boils at 101.70°C?

Answer 1

The molality of the solution is 3.32 mol/kg.

Explanation:

Water has a `K_b` of `"0.512 °C·kg·mol"^"-1"` of ions or molecules in solution.

This means that, for every mole of particles in 1 kg of water, the interference with the water molecules will raise the boiling point by 0.512 °C.

The normal boiling of water is 100.00 °C.

101.70 - 100.00 = 1.70, so the increase of the boiling point is 1.70 °C.

`ΔT_b = K_bm`

∴ `m =( ΔT_b)/K_b = "1.70 °C"/("0.512 °C·kg·mol"^"-1") = "3.32 mol·kg"^"-1"`

The molality is 3.32 m.

This molality could be achieved by dissolving 1 mol of a salt like `"CaCl"_2`, which ionizes into three particles, in 1 kg of water or by dissolving 3 mol of a compound like sugar in 1 kg of water.