$A$ $5 \%$ solution (by mass) of cane sugar in water has a freezing point of $271 \, K$. Calculate the freezing point of a $5 \%$ glucose solution in water,given that the freezing point of pure water is $273.15 \, K$.

(N/A) For cane sugar solution:
$\Delta T_{f} = (273.15 - 271) \, K = 2.15 \, K$.
$5 \%$ solution means $5 \, g$ of solute in $95 \, g$ of water.
Molar mass of cane sugar $(C_{12}H_{22}O_{11}) = 342 \, g \, mol^{-1}$.
Molality $(m) = \frac{5 / 342}{0.095} \, mol \, kg^{-1} = 0.1537 \, mol \, kg^{-1}$.
$K_{f} = \frac{\Delta T_{f}}{m} = \frac{2.15}{0.1537} = 13.99 \, K \, kg \, mol^{-1}$.
For glucose solution:
Molar mass of glucose $(C_{6}H_{12}O_{6}) = 180 \, g \, mol^{-1}$.
Molality $(m) = \frac{5 / 180}{0.095} \, mol \, kg^{-1} = 0.2926 \, mol \, kg^{-1}$.
$\Delta T_{f} = K_{f} \times m = 13.99 \times 0.2926 = 4.09 \, K$.
Freezing point of glucose solution $= 273.15 - 4.09 = 269.06 \, K$.