Question

the observed vapor pressure lowering of a CaCl2 solution is 5 Torr at 20C. Calculate the molality of the CaCl2 in the solution

Answer 1

Vapor pressure lowering, P = Xsolute Psolvent

Psolvent = Vapor pressure of pure water at 20 C = 17.5 torr

Xsolute = Mole fraction of solute (CaCl₂) = No. of moles of CaCl₂/Total no. of moles of solution

Molar mass of CaCl₂ = 110.98 g/mol

P = Xsolute Psolvent = 5 torr

Xsolute = 5 torr/17.5 torr = 0.286

Xsolute = 0.286 = No. of moles of CaCl₂/Total no. of moles of solution

Let us consider weight of solvent = 1kg = 1000 g

No. of moles of solvent water = wt/molar mass = 1000 g/18.01 g/mol = 55.5 moles

0.286 = No. of moles of CaCl₂/(No. of moles of CaCl₂ + 55.5 moles)

0.286 (n1 + 55.5) = n1

0.286 n1 + 15.873 = n1

0.286n1 -n1 = -15.873

0.714 x = 15.873

x = 15.873/0.714 = 22.23 moles

Molality of CaCl₂ = No. of moles of CaCl2 /Kg of solvent) = (22.23 moles/ 1 kg of solvent) = 22.23 m

or it can determined by direct formula :

molality = Xsolute x 1000 /(1-Xsolute) X Molar mass of solvent = 0.286 x 1000/(1-0.286) x 18.01 = 286 m/(0.714x18.01) = 286 m/12.86 = 2.23 m