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Question
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. The freezing point of 5% solution (by mass) of glucose in water is
Options
A.271 K
B.273.15 K
C.269.07
D.277.23 K
Solution
From the relation
ᐃTf = Kf ×
It is obvious that
ᐃTf ∝
∴
Cane sugar solution Glucose solution
ᐃTf1 = 273.15 = 271 ᐃTf2
m1 = 324 m2 = 180
Hence
ᐃTf2 = 4.085 K
So freezing point of glucose solution
= 273.15 - 4.085
= 269.05 K
ᐃTf = Kf ×
It is obvious that
ᐃTf ∝
∴
Cane sugar solution Glucose solution
ᐃTf1 = 273.15 = 271 ᐃTf2
m1 = 324 m2 = 180
Hence
ᐃTf2 = 4.085 K
So freezing point of glucose solution
= 273.15 - 4.085
= 269.05 K
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