Question

A student dissolves 50.0g of CaCl2 in 752 mL of water at a temperature of 25.0C and places it in a sealed 1.00L container. The final volume of this solution is 752mL. The density of water is 0.998g/mL. The Kf for water is 1.86C/m. The vapor pressure of water at 25.0C is 23.76mmHg. What is the vapor pressure of water in the headspace above the solution? What is the number of moles of water in the headspace above? what is the freezing point of the solution?

Answer 1

mass of water taken = d*v

                     = 0.998*752

                     = 750.5 g

no of mole of water = 750.5/18 = 41.7 mol

no of mole of CaCl2 = 50/111 = 0.45 mol

molfraction of CaCl2 = nCaCl2/ntotal

                      = 0.45/(41.7+0.45)

                      = 0.0107

P0-P/P0 = i*XCaCl2

i = vanthoffactor of CaCl2 = 3

(23.76-p)/p = 3*0.0107

P = vapor pressure of solution = 23.021 mmhg.

DTf = i*Kf*m

m = molality = w/mwt*1000/W in g

               = (50/111)*(1000/750.5)

               = 0.6 m
(T0-Tf) = i*Kf*m

(0-x) = 3*1.86*0.6

x = freezing point of solution = -3.348 c