# Exercise 7 | Colligative Properties of Solutions

One of the reason salt is added to water is because it increases the boiling point of the water, so it will cook better.

How much salt NaCl(s) would have to be dissolved in 1.00 L of water in order to raise the boiling point by 1°C?

Data: boiling point elevation constant of water = 0.513 K.kg.mol-1

The magnitude of the boiling point elevation ΔTb (in K) is:

ΔTb = i x Kb x m

ΔTb = Tb NaCl solution – Tb water

i = van’t Hoff i-factor = number of ions produced per formula unit

m = solution molality (in mol.kg-1)

Kb = boiling point elevation constant of water = 0.513 K.kg.mol-1

We want ΔTb = 1°C = 1 K

NaCl (s) + H2O (l) → Na+ (aq) + Cl- (aq)

⇒ i = 2

ΔTb = i x Kb x m

⇒ m =  = 0.975 mol.kg-1

m = $\frac{{\mathrm{n}}_{\mathrm{NaCl}}}{{\mathrm{m}}_{\mathrm{water}}}$ =

mNaCl = m x MNaCl x mwater = 0.975 x 58.44 x 1.00 = 57.0 g