A solution of glucose (Molar mass = 180 g mol–1) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water Kf and Kb are 1.86 K kg mol–1 and 0.512 K kg mol–1 respectively.
Given, boiling point of solution, Tb = 100.20°C = 373.35 K
Kf and Kb = 1.86 K kg mol–1 and 0.512 K kg mol–1 respectively.
We know, boiling point of pure water, Tb° = 100°C = 373.15 K
And freezing point of pure water, Tb°= 0°C = 273.15 K
Δ Tb = Kbm
Tb - Tb°= 0.512 
m
0.20 = 0.512 
m
m = 0.390 mol/kg
Now, putting the above value in Δ Tf = Kfm
Δ Tf = 1.86 
0.390
= 0.725 K
And, Δ Tf = Tf° - Tf
Tf = 273.15 – 0.725
= 272.425 K
Therefore, the freezing point of the same solution is 272.425K.
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