Question

1.The freezing point of an aqueous solution prepared by adding 0.0100 mol of acetic acid to 100. g of water is -0.190 C. The freezing point depression of pure water is 0.000 C, and the freezing point depression constant for water is 1.86 C/m. What is the value for the van't Hoff factor for acetic acid in the aqueous solution. You must show work to support your response.

2. Which of the following aqueous solutions should have the lowest freezing point: 0.0500 m C6H12O6, 0.0300 m KBr, or 0.0150m Na2SO4? You must show work to support your answer.

3. Arrange the following solutions in order of increasing freezing point depression: 0.10 m MgCl2 in water , i=2.7, Kf=1.86 C/m; 0.20 m toluene in diethyl ether, i=1.00, Kf=1.79 C/m; and 0.20 m ethylene glycol in ethanol, i=1.00, Kf=1.99 C/m. You must show work to support your answer.

Sigfigs!!!!!!!

Answer 1

1)

Tf = 0 - (0.190) = 0.190 oC

molality = moles of solute / mass of solvent (kg)

            =0.0100 / 100 x 10^-3

             = 0.1 m

Tf = i x Kf x m

0.190 = i x 1.86 x 0.1

van't Hoff factor for acetic acid = i = 1.02

2. )

freezing point inversly proportiaonal to the molality

0.0500 m C6H12O6 = i x m = 1 x 0.05 = 0.05 m

0.03 m KBr = i x m = 2 x 0.03 = 0.06 m

0.015 Na2SO4 = i x m = 3 x 0.015 = 0.045 m

least freezing point : KBr

3. )

MgCl2 freezing point depression = i x Kf x m = 2.7 x 1.86 x 0.10 = 0.502 oC

toluene in diethyl ether freezing point depression = 1 x 1.79 x 0.20 = 0.358 oC

ethylene glycol in ethanol = 1 x 1.99 x 0.20 = 0.398 oC

order :   MgCl2 > ethylene glycol in ethanol > toluene in diethyl ether