Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH)?

Question

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 11.2 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 1.00 x10^2 torr.

Answer 1

This will be an approximation, using Raoult's Law (the sum of the partial pressures is based upon the mole fraction or mole% of the mixture).

If the vapor pressure on Ethanol is 100 torr at 35° C and it is reduced to 100 - 11.2 = 88.8 torr.

1.00 kg of ethanol is 1000 g / 46.1 g/mol = 21.69 moles of Ethanol.

To reduce the pressure by 11.2 %, 88.8 mole% must be Ethanol and 11.2 mole% must be Ethylene glycol. If 21.69 moles is 88.8 mole%, then 1.00 mole% = 21.69 / 88.8 = 0.244.

11.2 mole% must be 11.2 * 0.224 = 2.51 moles. Therefore 2.51 moles of Ethylene glycol must be added. One mole = 62.1 g/mole.

2.51 moles * 62.1 g/mol = 155.87 grams or about 156 grams of Ethylene glycol.