Question

Pure ethanol has a boiling point of 78.0 °C. What is the boiling point of a solution that contains 12.7 grams of KI (MW = 166 g/mol) dissolved in 125 grams of ethanol (MW = 46.0 g/mol)? Assume that KI is a strong electrolyte in ethanol. (Ethanol: kb= 1.22 °C/m)

Answer 1

ing point g point we know STO=k6om Where Atb = elevation in boiling poln = To-Tb I to a boiling of solution (KI + Ethand) Th= Boiling point of pure ethanol = 374 K6= Constant = 1022 kg C./m. - 780€ / m a molality of solution = number of moles of KI Weight of ethandin kg = 12.7/166 m - 12.7 xivoo maha 125/1000 kg- 1668 125 ng

To To - 78= 1.22 x 12.78 1000 oC 1664 125 or To' = 78+ 0.75 oC or to a 78.75 °C elivation in boilina point of solution of ethemos en KI woultbe 78.75€ ef KI is not behave as strong electrolyte. AS KI behave as strong electrolyte it will dissociate as kt+1- l.e. from one KI ,, eve will get 2 ions and as elivation of boiling point is a colligative property, so it wile changes. So the equation (new) for elivation of a boiling point is To - Tb = 2xKbx m or To = 2x 1.22 x 12.7 x 1000 +78 cc T66 X125 = 78+1:49 OL = 79.49°C So the esikation in boiling point of Solution OG KI in ethanol is 79.49 €