Question

An aqueous solution of monobasic acid .1 g in 21.7 g of water freeses at 272.813 if kf value is 1.86 then molecular weight of solute

Answer 1

(Delta)Tf = Kf × m

Let molecular mass be M

272.813 = 1.86 × m

Where m is molality,

m = (0.1/M) / (21.7/1000)

272.813 = 1.86 × 0.1/ M × 1000/21.7

On solving,

M = 0.0314 kg or 31.4 g (ans.)