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NCERT Class XII Chemistry
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Question : 20
Total: 53
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure water is 273.15 K.
Solution:  
Molality of sugar solution =
5
342
×
1000
100
=0.146 m
ΔTf for sugar solution =273.15−271=2.15
ΔTf=Kf×m‌‌∴Kf=
ΔTf
m
=
2.15
0.146

Molality of glucose solution =
5
180
×
1000
100
=0.278 m
∴‌‌ΔTf (Glucose) =
2.15
0.146
×0.278=4.090
∴‌‌ Freezing point of glucose solution =273.15−4.09=269.06 K
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