Question

a solution is prepared by dissolving 27.8 grams of lithium sulfate (Li2SO4) in 500grams of water. caclulate boiling point of the solution.

Answer 1

Solution: given problem is based on colligative property elevation in boiling point.

∆Tb = i × kB × m

∆Tb is elevation in boiling point , i is vanthoff factor for Solute ( Li ₂SO₄) = 3

Kb is boiling constant for solvent (water) = 0.512 °C / m

m is molality = number of moles of solute present in 1 kg of solvent

Number of moles of lithium sulphate = mass ÷ molarmass = 27.8 g ÷ 110 g/mol = 0.2527 mol

Mass of water = 500 g = 0.5 kg

Molality (m) = 0.2527 mol ÷ 0.5 kg = 0.5054 m

Now , ∆Tb = 3 × 0.52 °C/m × 0.5054 m = 0.788 °C

So elevation in boiling point = 0.788 °C

Boiling point of water at 1 atm = 100 °C or 373.15 K

Then, boiling point of solution ( lithium sulphate in water ) = boiling point of water + ∆Tb = 100 + 0.7884 = 100.788 °C

Boiling point of solution = 100.788 °C or 373.938 K