Given:

Sucrose solution =4%(w∕w)
M=342gmol−1
Freezing point of solution =271.15K
Freezing point of pure water =273.15K
Glucose solution =5%
M=180gmol−1
To calculate:

Freezing point of 5% glucose solution.

Formula:

∆Tf‌=i×Kf×m
m‌=‌

‌ moles of solute ‌

Kg‌ of solvent ‌

‌ moles ‌‌=‌

‌ mass ‌

‌ molar mass ‌

Sucrose solution is 4%(w∕w) which means there is 4.0 grams of sucrose dissolved in 100g of solution.
Mass of solution = mass of solute + mass of solvent 100.0g=4.0g+ mass of solvent.
Hence, mass of solvent =100.0−4.0=96.0g
96g×‌

1‌kg

1000g

=0.096‌kg
Moles of solute
‌ moles ‌‌=‌

‌ mass ‌

‌ molar mass ‌

=‌

4.0g

342gmol−1

‌=0.011695‌ moles ‌
Molality of solution:
m=‌

‌ moles of solute ‌

kg‌ of solvent ‌

=‌

0.011695‌moles

0.096‌kg‌ water ‌

=0.1218m
Sucrose is a non-electrolyte, hence i=1
∆Tf‌=‌ Freezing point of solvent ‌
‌−‌ Freezing point of solution ‌
∆Tf‌=273.15K−271.15K=2.00K
∆Tf‌=i×Kf×m
‌2.00K=i×Kf×0.1218
‌Kf=16.42Km−1
For glucose solution,
Glucose solution is 5%(w∕w) which means there is 5.0 grams of glucose dissolved in 100g of solution.
Mass of solution = mass of solute + mass of solvent 100.0g=5.0g+ mass of solvent.
Hence, mass of solvent =100.0−5.0=95.0g
95.0g×‌

1‌kg

1000g

=0.095‌kg
Moles of solute
‌ moles ‌=‌

‌ mass ‌

‌ molar mass ‌

=‌

5.0g

180gmol−1

=0.0277‌ moles ‌
Molality of solution:
m=‌

‌ moles of solute ‌

kg‌ of solvent ‌

=‌

0.0277‌ moles ‌

0.095‌kg‌ water ‌

=0.2923m
Glucose is a non - electrolyte, hence i=1
∆Tf=i×Kf×m
∆Tf‌=1×16.42×0.2923
∆Tf‌=4.801Km−1
∆Tf‌=‌ Freezing point of solvent ‌
‌−‌ Freezing point of solution ‌
4.801‌=273.15K−‌ Freezing point of solution ‌
‌ Freezing point of solution ‌‌=273.15K−4.801K
‌=268.35K
Thus, the freezing point of the glucose solution is 268.35K.