Given:

Sucrose solution $=4\%\left(\mathrm{w}∕\mathrm{w}\right)$
$\mathrm{M}=342\mathrm{g}{\mathrm{mol}}^{-1}$
Freezing point of solution $=271.15\mathrm{K}$
Freezing point of pure water $=273.15\mathrm{K}$
Glucose solution $=5\%$
$\mathrm{M}=180\mathrm{g}{\mathrm{mol}}^{-1}$
To calculate:

Freezing point of $5\%$ glucose solution.

Formula:

$∆{\mathrm{T}}_{\mathit{f}}=\mathit{i}\times {\mathrm{K}}_{\mathit{f}}\times \mathit{m}$


Sucrose solution is $4\%\left(\mathrm{w}∕\mathrm{w}\right)$ which means there is 4.0 grams of sucrose dissolved in $100\mathrm{g}$ of solution.
Mass of solution $=$ mass of solute + mass of solvent $100.0\mathrm{g}=4.0\mathrm{g}+$ mass of solvent.
Hence, mass of solvent $=100.0-4.0=96.0\mathrm{g}$
$96\mathrm{g}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}=0.096\mathrm{kg}$
Moles of solute


Molality of solution:

Sucrose is a non-electrolyte, hence $\mathrm{i}=1$


$∆{\mathrm{T}}_{\mathit{f}}=273.15\mathrm{K}-271.15\mathrm{K}=2.00\mathrm{K}$
$∆{\mathrm{T}}_{\mathit{f}}=\mathit{i}\times {\mathrm{K}}_{\mathit{f}}\times \mathit{m}$
$2.00\mathrm{K}=\mathit{i}\times {\mathrm{K}}_{\mathit{f}}\times 0.1218$
${\mathrm{K}}_{\mathit{f}}=16.42{\mathrm{Km}}^{-1}$
For glucose solution,
Glucose solution is $5\%\left(\mathrm{w}∕\mathrm{w}\right)$ which means there is 5.0 grams of glucose dissolved in $100\mathrm{g}$ of solution.
Mass of solution $=$ mass of solute + mass of solvent $100.0\mathrm{g}=5.0\mathrm{g}+$ mass of solvent.
Hence, mass of solvent $=100.0-5.0=95.0\mathrm{g}$
$95.0\mathrm{g}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}=0.095\mathrm{kg}$
Moles of solute

Molality of solution:

Glucose is a non - electrolyte, hence $\mathrm{i}=1$
$∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$
$∆{\mathrm{T}}_{\mathrm{f}}=1\times 16.42\times 0.2923$
$∆{\mathrm{T}}_{\mathrm{f}}=4.801{\mathrm{Km}}^{-1}$




$=268.35\mathrm{K}$
Thus, the freezing point of the glucose solution is $268.35\mathrm{K}$.