Question

An aqueous CaCl2 solution, which has a mole fraction XCaCl2 = 0.0016 has a freezing point of -0.35 ºC. What is the actual value of the van't Hoff factor (i) for CaCl2 in this solution?

Answer 1

Date dre The data given in the question. mole braction ols Xcach = 0.0016 The breezing point of the solution ich = -0.350 The molal depression constant of tho (solvents ÿ 1.86°c kg mol (TG)= Freezing point of pure tho = 06. To = freezing point of solution = -0.350 Hence of depression in freezing point Tf - To o-(-0.356 ) = = 0.35€ From the collegative properties we know sto = iKfm where Kf = molal depression Constant of tho=1.86°c Kg molet m = molality of the solution i=vant Hoff tactos can find molality of aqueous solution from more traction solute cach 0.0016 The mole traction cach = 0.0016 indicate that it total moles of solute cach and solvent the mae then cach present ý 0.0016 mole Hence moles of to solvent-1-0.0016] Hence molality - moles of solute mass of forventinky 2 Now we X each 1 one + 0.9984

0.0016 mol molality of Cach solution 0.9989 18x16 x 18xlokg 0.089 m (Note molar mass of tho-18 gf mole) Hences molality of cach aqueous solution - 0.0889 mol kgl. Now BTG - i kfx m => i= ATG 0.35c Krom 1.86% Kg. more to 089 t = 2.114 mot kgl There cach is an electrolyte, hence dissociate to cart and sé so its vant Hoff factor y always greater than i Hence the actual value of Vant - Hoff bactor of cach in this saution y 2.114.