Question

You add 1.3 kg of ethylene glycol (C2H6O2) antifreeze to 4692 g of water in your car's radiator. What is the boiling point of the solution? The Kb for water is 0.512 °C/m. Enter to 2 decimal places.

Answer 1

ethylene glycol mass = 1.3 kg = 1300 g

moles of glycol = mass / ( molar mass of glycol) = ( 1300 g) / ( 62g/mol) = 20.97 mol

water mass = 4692 g = 4.692 kg

molality = moles of glycol / mass of water in kg = ( 20.97 mol) / ( 4.692 kg) = 4.47 mol

dT = i x Kb x m   is the relation , where dT = change in BP , i = vantoff factor = 1 for non dissociable sollutes

Kb = 0.512 C/m ,   = molality = 4.47 , we find dT

dT = 1 x 0.512 C/m x 4.47 m

= 2.29 C = Tb ( solution) - Tb ( solvent)

2.29C = Tb (solution) - 100

Tb (solution) = 102.29 C