# A 4% Solution(W/W) of Sucrose (M = 342 G Mol–1) in Water Has a Freezing Point of 271.15 K. Calculate the Freezing Point of 5% Glucose (M = 180 G Mol–1) in Water. - Chemistry

#### Question

A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water.
(Given: Freezing point of pure water = 273.15 K)

#### Solution

Molality (m) = "n"/"W"_("solvent"("kg"))

For sucrose solution:

m =(4/342)/(96/1000)= 4/342xx1000/96 = 0.122"m"

(Δ Tf)1 = (273.15 − 271.15)K = 2K

(Δ Tf)1 = Kfm = Kf × 0.122

2= Kf × 0.122    .......(1)

For glucose solution :

m=(5/180)/(95/1000)=5/180xx1000/95= 0.292"m"

(Δ Tf)2 =Kf × 0.292  .......(2)

Dividing eqn. (2) by (1)

(Delta "T"_"f")_2/2 = ("K"_"f"xx0.292)/("K"_"f"xx0.122

(Delta"T"_"f")_2 = 0.292/0.122xx2 = 4.79

Tf = 273.15 − 4.79 = 268.36 K

Freezing point of glucose solution will be 268.36 K

Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point
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