Question

0.5 g KCl was dissolved in 100g of water and the solution freezes at - 0.24 C. Calculate the percentage ionisation of the salt. Kf for water is 1.86 C

Answer 1

Given Info : 0.5g of KCl was dissolved in 100g of water and the solution freezes at -0.24°C. Kf of water is 1.86 K kg/mol

To find : the percentage ionisation of the salt is...

solution : no of moles of KCl = mass of KCl/molar mass of KCl

= 0.5g/74.5 g/mol

= 0.0067 mol

mass of solvent (water) = 100g

molality , m = no of moles of KCl/mass of solvent in kg

= 0.0067 mol/(100/1000 kg)

= 0.067 mol/kg

dissociation of KCl is .

KCl ⇒ K⁺ + Cl¯

1 - α α α

so, i = 1 - α + α + α = 1 + α

using formula, ∆Tf = i × Kf × m

⇒0.24 = (1 + α) × 1.86 × 0.067

⇒0.24/(1.86 × 0.067) = 1 + α

⇒1.926 = 1 + α

⇒ α = 0.926

Therefore the percentage ionisation of the salt is 92.6 %