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

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \frac{\mathrm{w}}{\mathrm{m}}\times \frac{1000}{\mathrm{W}}\mathrm{and} \frac{∆{\mathrm{T}}_{{\mathrm{f}}_2}}{∆{\mathrm{T}}_{{\mathrm{f}}_1}}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}$
Here, ${\mathrm{m}}_1(\mathrm{cane} \mathrm{sugar},{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11})=342 \mathrm{g} {\mathrm{mol}}^{-1}{\mathrm{m}}_2(\mathrm{gluocse},{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6)=180 \mathrm{g} {\mathrm{mol}}^{-1}$
                      $$\begin{gathered} ∆{\mathrm{T}}_{{\mathrm{f}}_1}=273.15-271=2.15 \mathrm{K} \\ \frac{∆{\mathrm{T}}_{{\mathrm{f}}_2}}{2.15}=\frac{342}{180}\Rightarrow ∆{\mathrm{T}}_{{\mathrm{f}}_2}=4.085 \mathrm{K} \end{gathered}$$
So, freezing point of glucose in water  $=273.15-4.085=269.07 \mathrm{K}$