Question

6.   What is the freezing point of a solution (water) that boils at 105.0^oC??

7. The Henry's law constant of helium gas in water at 30^oC is 3.7 X 10-4 M/atm and the constant for Nitrogen (N₂) at 30^oC is 6.0 X10-4 M/atm

          a. How many M of helium are dissolved in a solution if it is at 3 atm?

          b. How many M of Nitrogen are dissolved in a solution if it is at 75 atm?

8.   List the following aqueous solutions in order of decreasing freezing point:

0.04 m glycerine                 0.020 m Kbr                     0.03 m Phenol

Answer 1

1. use equation for boiling point elevation
?Tb = Kb x m x i

and this equation for freezing point depression
?Tf = Kf x m x i

Kb = ebullioscopic constant (aka boiling point elevation constant) for the solvent (water in this case) = 0.512 C/m

Kf = cryoscopic constant (aka freezing point depression constant) for the solvent. = 1.86 C / m for water

m = molality of the solution = moles solute / kg solvent
i = van't hoff factor = # ions produced in solution / molecule solvent

got that?... the idea here is m and i are constant for the solution. the concentration doesn't change. neither does the amount of ions in solution.

so from the equations above...

?Tb / Kb = m x i = ?Tf / Kf

rearranging
?Tf = ?Tb x Kf / Kb

since the normal boiling point of water = 100C

?Tf = (105.0 C - 100 C) x (1.86 C/m) / (0.512 C/m) = 18.16 C

since the normal freezing point of water = 0C and ?Tf = 18.16C, the depressed freezing point = -18.1 C

2.

p*K = c;
so for He it's: 3.7*10-4*1.4M = 0.518mmol/l

for N2: .84mmol/l.

3. delta T = kf x m x i
i is the number of ions dissolved in the solution if the solute is an electrolyte. In this case only KBr is an electrolyte and i = 2
1) delta T = 0.04 x 1.86 =0.0744
2) delta T = 0.020 x 1.86 x 2 = 0.0744
3) delta T = 0.030 x 1.86 = 0.0558