Q15 of 26 Page 1

Calculate the freezing point of solution when 2 g of Na2SO4 (M = 142 gmol-1) was dissolved in 50 g of water, assuming Na2SO4 undergoes complete ionization. (Kf for water = 1.86 K kg mol-1)

Given, Mass of solvent (w1) = 50 g


Mass of solute (w2) = 2 g


Molar mass of solute (M2) = 142 g/mol


Kf = 1.86 K kg mol-1


Na2SO4 → 2Na+ + SO42-


i(no. of particles after dissociation) = 3


ΔTf =


= = 1.56 K


Now, Tfo for water i.e freezing point of water = 273.15 K


ΔTf = Tf° - Tf


1.56 = 273.15 – Tf


Tf = 271.59 K


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