Question

An ethylene glycol solution contains 24.4 g of ethylene glycol (C2H6O2) in 91.8 mL of water. (Assume a density of 1.00 g/mL for water.)

Determine the boiling point of the solution.

Express you answer in degrees Celsius.

Answer 1

ethylene glycol is the non electrolyte it doesn't undergo association or dissociation

mass of water = volume * density = 91.8 mL x 1.00 g/mL =91.8 g = 0.0918 Kg

moles C₂H₆O₂ = given mass / molar mass= 24.4 g /62.068 g/mol=0.3931mol

molality = no of moles of solute / mass of solvent in (kg) = 0.3931mol/ 0.0918Kg = 4.2821 m

T = molality * K_b= 4.2821 * 0.512= 2.1924 C⁰   Kb for etylene glycol = 0.512 C⁰

boiling point_sol = boiling point of water +T = 102.1924 °C