∆Tt=(273.15−269.15)K=4K
Molar mass of sucrose (C12H22O11)
‌=(12×12)+(22×1)+(11×16)
‌=342gmol−1
10% solution of sucrose in water means 10g of sucrose is present in (100−10)g of water.
‌ Number of moles of sucrose ‌=‌‌

10

342

‌=0.0292‌mol
Therefore, molality of the solution
=‌

0.0292×1000

90

=0.3244‌mol‌kg−1
We know that ∆Tt=Kf×m
⇒Kf=‌

∆Tt

m

=‌

4

0.3244

=12.33K‌Kg‌mol−1
Molar mass of glucose (C6H12O6)
=(6×12)+(12×1)+(6×16) =180gmol−1
10% solution of glucose in water means 10g of glucose is present in (100−10)g of water.
‌ Number of moles of glucose ‌ =‌

10

180

=0.0555‌mol
Therefore,molality of the solution
‌=‌

0.0555×1000

90

‌=0.6166‌mol‌kg−1
‌ We know that ‌∆Tt‌=Kf×m
⇒‌‌∆Tt‌=12.33×0.6166=7.60K
So, the freezing point of 10% glucose solution in water is (273.15−7.60)K=265.55K