Question

*22. Calculate the freezing point and boiling point of the next solution: 0.35m glycerol (non-electrolyte) Kf-1.990C/m; TfO s - 114.6C; Kb0 x 1.220C/m; Tb0 x 78.4C.

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Answer 1

Answer: Freezing point of solution - If Tj - off Where DTg = depression in freezing point off = Kfx m m= molality € 0.35m here) so Tf - freezing point of pure solvent - - 114.6² so If a Te - Atf = Tq - Kfm Tea - 114.6°C - 1.99°C X 0.35 m If = -11 5.3c/ < freezing point of 235m solution Boiling point of solution - Tb = Tot sto where AT6 = Elevation in boiling point ATG = Kg x m where m= molality of solute / solution (= 0.35m here) 1 ho= Boiling potht elevation constant = 1.22°C/m): so The 78.40 + 1.22 c / mp x 0.35 h To = 78.827°C It Boiling point of 0.35m solution.