SolutionHard
Question
The freezing point of aqueous solution that contains 3% urea, 7.45% KCl and 9% of glucose is (given Kf of water = 1.86 and asume molarity = molality).
Options
A.290 K
B.285.5 K
C.267.42 K
D.250 K
Solution
ᐃTf = i.m. Kf
ᐃTf = i1m1Kf + i2 m2 Kf + i3 m3 Kf = (m1 + 2m2 + m3) Kf
ᐃTf =
× 1000 × 1.86
ᐃTf = 3 × 1.86 = 5.58
Tf of solution = 273 - 5.58 = 267.42 K Ans.
ᐃTf = i1m1Kf + i2 m2 Kf + i3 m3 Kf = (m1 + 2m2 + m3) Kf
ᐃTf =
× 1000 × 1.86 ᐃTf = 3 × 1.86 = 5.58
Tf of solution = 273 - 5.58 = 267.42 K Ans.
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