Question

a) An ethylene glycol solution contains 25.2 gg of ethylene glycol (C2H6O2)(C2H6O2) in 96.8 mLmL of water. (Assume a density of 1.00 g/mLg/mL for water.)

i) determine freezing point
ii) determind boiling point

degrees C

Answer 1

Arswer'- fuiven : mass of eHaylene glycol = 25:29 volume of water = 968 mL Deusity of Water = 1.60 g/ml Freezing point = (OTA) = ? Boilling poput - (Oto) = ? To calculate freezing point & Brilling point we need Molality of ethylene glycol Mass of ethylenglycol Moles of ethylene glycol = Molar Mass of ethylene glycol. = 25.29 62.068 31mole Moles of ethylene glycol. = 0.4060 moles Mass of water = volume x density = 96.8 x 1.00 = 96.8 g Mass of water = 0.0968 kg Molality of Ethylene glycol a Moles of ethylene glycol Mass of water in kg 0-4060 0.0968 Molality of Ethylene glyat=4.194 m] o freezing point atf = xz x molality = 1.86X4.194 BTA = 7.8008°C Freezing point of solution = freezing point of water - ATE -0.00 - 7.8008°C Freezing point of solution = -7.8008°C В) ДТ) = K, хто ду = 0.512 X 4.194 - DTb = 2.1473°C Scannedilling point of solutio= B.P. of Walest Tb = 102.1473 100+ 9.1473 CamScoilling point of solution = 102.1473°C]