Mass of glucose, WB = 3.6 g
Molecular mass of glucose = 180 g mol-1
Mass of solvent, WA = 50 g
Kf for H2O = 1.86 K/m
= 1.86 K kg-1 mol-1
MB = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,\bigtriangleup T_f}\)
∴ \(\bigtriangleup T_f\) = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,M_B}\)
= \(\frac{1000\,g\,kg^{-1}\,\times\,1.86\,K\,kg^{-1}\,mol^{-1}\,\times\,3.6\,g}{50\,g\times180\,g\,mol^{-1}}\)
= 0.744 K.
i.e., ΔTf = T°f – Tf
= 0.744 K
∴ Freezing point of the solution,
Tf = T°f – ΔT
= 273 K – 0.744 K
= 272.3 K