If the osmotic pressure of a 4.26 times 10^-2-M aqueous solution of K_2SO_4 was

Question

If the osmotic pressure of a 4.26 times 10^-2-M aqueous solution of K_2SO_4 was found to be 2.93 atm at 20 degree C, what would be the "observed" van't Hoff factor? cf Zumdahl "Chemical Principles" 7th ed., sec17.6 "Osmotic Pressure" and sec17.7 "Colligative Properties of Electrolyte Solutions", pp871-876. Use the above van't Hoff factor to predict the freezing point of this solution. that the corresponding molality would be 4.30 times 10^-2 m, given that the density of the solution is 0.997 g/cm^3. degree C cf Zumdahl "Chemical Principles" 7th ed., sec17.7 "Colligative Properties of Electrolyte Solutions", pp875-6. Use the above van't Hoff factor to predict by how many degrees the boiling point of this solution will be elevated above that for pure water. degree C

Explanation / Answer

1) osmotic pressure (Pi) = iCRT

osmotic pressure (Pi) = 2.93 atm

i = vanthoff factor = ?

c = concentration of solution = 0.0426 M

R = solution constant = 0.0821 l.atm.k-1.mol-1

T = 293 k

2.93 = x*0.0426*0.0821*293

i = vanthoff factor = 2.86

2)

DTf = i*Kf*m

i = vanthoff factor of K2SO4 = 2.86

Kf of water = 1.86 c*kg/mol

m = molality of solution = (wt of solute/Mwt)*(1000/wt of solvent in grams)

   = 0.043

DTf = Twater - Tsolution

    = 0 - x

0 - x = 2.86*1.86*0.043

x = freezingpoint of solution = -0.23 C

3)

DTb = i*Kb*m

DTb = boilingpoint of solution - boiling point of solvent = ?

i = vanthoff factor = 2.86

Kb of water = 0.512 c*kg/mol

m = molality of solution = (wt of solute/Mwt of solute)*(1000/wt of solvent in gr)

= 0.043 m

x = 2.86*0.512*0.043

= 0.063 C

no of degrees of boilingpoint elevated = 0.063 C

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