19. Calculate the osmotic pressure of a solution containing 24.9g of glycerin (C

Question

19. Calculate the osmotic pressure of a solution containing 24.9g of glycerin (C3H8O3) in 255.0mL of solution at 290K

20. A solution containing 29.30mg of an unknown protein per 21.5mL solution was found to have an osmotic pressure of 3.40torr at 38?C. What is the molar mass of the protein?

21. Calculate the freezing point of the following solutions, assuming complete dissociation.

A. 0.123m K2S

B. 22.1g of CuCl2 in 477g water

C. 5.9% NaNO3 by mass (in water)

D. Calculate the boiling point of the solution in part A, assuming complete dissociation.

E. Calculate the boiling point of the solution in part B, assuming complete dissociation.

F. Calculate the boiling point of the solution in part C, assuming complete dissociation.

22. Use the van't Hoff factors in the table to compute each of the following.

Solute i Measured

FeCl3 3.4

K2SO4 2.6

MgCl2 2.7

A. The melting point of a 0.120m iron(III) chloride solution.

B. The osmotic pressure of a 7.9

Explanation / Answer

osmotic Pressure = CRT

Moles of Glycerin = 24.9 / 92 = 0.271

Molarity = 0.271 / 0.255 = 1.061

Osmotic Pressure = 1.061 x 0.082 x 290 = 25.24 atm.

20)

3.4 torr = 0.00447 atm

0.00447 = C x 0.082 x 311

=> C = 1.75 x 10^-4 M

=> n / 0.0215 = 1.75 x 10^-4

=> n = 3.77 x 10^-6 moles = 0.0293 / M

=> M = 7768.5

21)

A) delta Tf = 1.86 x 0.123 x 3 = 0.686 C

=> Tf = - 0.686 degree C

B) m = (22.1 / 134.45) / 0.477 = 0.345

Tf = - 1.86 x 0.345 x 3 = -1.923 degree C

C) m = (5.9 / 85) / 0.0941 = 0.738

Tf = - 1.86 x 0.738 x 2= -2.744 degree C

D) Tb = 100 + (0.512 x 0.123 x 3) = 100.19 degree C

E) Tb = 100 + (0.512 x 0.345 x 3) = 100.53 degree C

F) Tb = 100 + (0.512 x 0.738 x 2) = 100.756

22)

A) Tf = - 0.12 x 1.86 x 3.4 = -0.759 degree C

B) OP = 0.079 x 0.082 x 320 x 2.6 = 5.39 atm

C) m = (1.4/ 95.2) / 0.0986 = 0.149

Tb = 100 + (0.512 x 2.7 x 0.149) = 100.206 degree C

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