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.
Create a free account to view solution
View Solution FreeMore Solution Questions
Among the following mixtures, dipole-dipole as the major interaction, is present in...The vapour pressure of benzene at 30oC is 121.8 mm. By adding 15g of non-volatile solute in 250 g of benzene, its molecu...Phenol dimerizes in benzene. If the observed molecular mass of phenol in solution is 120, then its degree of dimerizatio...Of the following measurements, the one most suitable for the determination of the molecular mass of oxyhaemoglobin, a mo...The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, then ...