Molar mass of cane sugar

C_{12}H_{22}O_{11} = 342 g mol^{-1}

Molality of sugar = \(\frac{5\times1000}{342\times 100}\)= 0.146

∆T_{2} for sugar solution = 273.15 – 271 = 2.15°

∆T_{f} = K_{f} x m

K_{f} = \(\frac{2.15}{0.146}\)

Molality of glucose solution = \(\frac{5}{180}\) x \(\frac{1000}{100}\)= 0.278 m

∆T_{f} (Glucose) = \(\frac{2.15}{0.146}\) x 0.278 = 4.09°K

Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K