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

For both the solutions W_B and W_A are the same
$\begin{array}{l}\frac{{\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)}_{\mathrm{g}}}{{\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)}_{\mathrm{s}}}=\frac{{\mathrm{M}}_{\text{sucrose }}}{{\mathrm{M}}_{\text{glucose }}}\\ {\mathrm{\Delta T}}_{\mathrm{f}\left(\mathrm{g}\right)}=\frac{342 \mathrm{g}/\mathrm{mol}}{180 \mathrm{g}/\mathrm{mol}}\times 2.15\mathrm{K}=4.085\mathrm{K}\end{array}$
freezing point of glucose solution = (273.15-4.085)

                                                      = 269.07K