Here, ΔT_f = (273.15 - 271) K

= 2.15 K

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 22 × 1 + 11 × 16

= 342 g mol^-1

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g

= 95 g of water.

Now, number of moles of cane sugar = \(\frac{5}{342}\)mol

= 0.0146 mol

Therefore, molality of the solution, m = \(\frac{0.0146\ mol}{0.095\ kg}\)

= 0.1537 mol kg^-1

Applying the relation,

ΔT_f = K_f × m

⇒ K_f = \(\frac{ΔT_f}{m}\)

= \(\frac{2.15\ K}{0.1537\ mol\;kg^{-1}}\)

= 13.99 K kg mol - 1

Molar of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol^-1

5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.

Number of moles of glucose = \(\frac{5}{180}\)mol

= 0.0278 mol

Therefore, molality of the solution,

= 0.2926 mol kg^-1

Applying the relation,

ΔT_f = K_f × m

= 13.99 K kg mol^-1 × 0.2926 mol kg^-1

= 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K = 269.06 K.