A 5% solution (by mass) of cane sugar in water has freezing point of 271 K.

Here, ΔTf = (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 $=\frac{5}{342} \mathrm{~mol}$

= 0.0146 mol

Therefore, molality of the solution, $m=\frac{0.0146 \mathrm{~mol}}{0.095 \mathrm{~kg}}$

= 0.1537 mol kg−1

Applying the relation,

ΔTf = Kf × m

$\Rightarrow K_{f}=\frac{\Delta T_{f}}{m}$

$=\frac{2.15 \mathrm{~K}}{0.1537 \mathrm{~mol} \mathrm{~kg}^{-1}}$

= 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.

$\therefore$ Number of moles of glucose $=\frac{5}{180} \mathrm{~mol}$

= 0.0278 mol

Therefore, molality of the solution, $m=\frac{0.0278 \mathrm{~mol}}{0.095 \mathrm{~kg}}$

= 0.2926 mol kg−1

Applying the relation,

ΔTf = Kf × 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.