Question

An aqueous solution of calcium chloride (CaCl2: MM=111.1g/mole) is 14% CaCl2 by mass. It has a density of 1.17g/ml.

a> determine the molality.

b> determine the freeing point of the solution (kf= 1.86 C/m)

c> determine the molarity of the solution.

d> how would the vapor pressure of this solution compare to that of pure water. Explain

Answer 1

molality = mol of S / kg solvent

assume: 100 g of solution

then

mass of CaCl2 = 100*0.14 = 14 g of CaCl2

mol of CaCl2 = masS/MW = 14/111.1 = 0.1260

kg of solvnet = (100-14)/1000 = 0.086

molal = 0.1260/0.086 = 1.4651 molal

b)

dTf = -Kf *m *i

i = 1+2 = 3 ions in solution

dtf = -1.86*1.4651 *3 = -8.175258

Tf = -8.175258 °C

d)

M = mol/V

mol = 0.1260

V = m/D = 100/1.17 = 85.470 ml

M = 0.1260 / ( 85.470/1000) = 1.474

d)

P° = Psolvent-xsolute*P°solvent

xsolute = mol S / (mol S + mol Water)

mol water = mass/MW = 86/18 = 4.777777

x = 0.1260 /(4.777777+0.1260 ) =0.0257

for water

assume T = 25°C

23.8 torr

then

P° = 23.8 -0.0257*23.8 = 23.18834 torr