Question

A solution is made by dissolving 2.3 moles of sodium chloride (NaCl) in 0.155 kilograms 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 the steps taken to solve this problem.

Answer 1

Answer:

115 °C

Explanation:

Data:

n(NaCl) = 2.3 mol

m(H₂O) = 0.155 kg

       Kb = 0.51 °C·kg·mol⁻¹

Calculations:

(a) Molal concentration

The formula for molal concentration (b) is

b = moles of solute/kilograms of solvent

b = 2.3/0.155

   = 14.8 mol/kg

(b) Boiling point elevation

The formula for the boiling point elevation ΔTb is

ΔTb = iKb·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For NaCl,

   NaCl(s)   ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol NaCl ⟶  2 mol particles     i = 1

ΔTf = 2 × 0.51 × 14.8

        = 15.1 °C

(c) Boiling point

Tb = Tb° + ΔTb

     = 100.0 °C + 15.1 °C

     = 115 °C