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
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