Question

The liquid in automobile cooling systems is prepared by dissolving ethylene glycol (HOCH_2CH_2OH) in water. Ethylene glycol has a molar mass of 62. 07 g/mol and a density 1. 115 g/ml at 50. 0 degree C. Calculate the vapor pressure at 50 degree C of a coolant solution that is 51. 0: 49. 0 ethylene glycol to water by volume At 50. 0degree C the density of water is 0. 9880 g/ml, and its vapor pressure is 92 torr. The vapor pressure of ethylene glycol is less than 1 torr at 50. 0 degree C.

Answer 1

Solution:

1) The vapor pressure of solvent is related to the mole fraction of the solvent and its pure vapor pressure:

vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent

2) Calculations

mass of ethylene glycol = density*volume = 1.115*51 = 56.865 g

moles of ethylene glycol --->56.865 g / 62.07 g/mol = 0.334 mol

Mass of water = 49*0.988 = 48.412 g

moles of water ---> 48.412 g / 18.0 g/mol = 2.69 mol

mole fraction of ethylene glycol = moles of ethylene glycol/total moles

mole fraction of ethylene glycol ---> 0.916 / (0.916+ 2.69) = 0.254

vapor pressure of solution = mole fraction * pure vapor pressure

vapor pressure of solution ---> 1torr * 0.254 = 0.254. torr