A 5% solution (by mass) of cane sugar in water has freezing point = 271 K and freezing point of pure water rs 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is

∆Tf=Kf×wm×1000Wand ∆Tf2∆Tf1=m1m2Here, m1(cane sugar,C12H22O11)=342 g mol-1m2(gluocse,C6H12O6)=180 g mol-1                      ∆Tf1=273.15-271=2.15 K ∆Tf22.15=342180⇒∆Tf2=4.085 KSo, freezing point of glucose in water  =273.15-4.085=269.07 K