Question

1. The liquid used in automobile cooling systems is prepared by dissolving ethylene glycol (HOCH2CH2OH) in water. Ethylene glycol has a molar mass of 62.07 g/mol and a density of 1.115 g/mL at 50.0°C.
Calculate the vapor pressure at 50°C of a coolant solution that is 55.0:45.0 ethylene glycol-to-water by volume. At 50.0°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°C. answer in torr

Answer 1

The vapor pressure of ethylene glycol is 1 torr at 50^oC, which is negligible in comparison to water. Thus, the solution contains a non-volatile ethylene glycol in a volatile solvent which is water. The partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. In symbols pi=χi p∘I where;

pi= the partial pressure  of component i

χi=The mole   fraction of component i

p∘i= the vapor pressure of the pure component

Assume that we have 100 ml of the solution.

Then we have 52 ml of EG and 48 ml of water.                                                       

Step 1. Calculate the moles of water

$Mass of water=48ml* 0.9880g /ml=47.42g$ Mass of water=48ml* 0.9880g /ml=47.42g

moles of water =47.42/18.02 =2.363

Step 2. Calculate the moles of Ethylene Glycol (EG)

Mass of EG=52.0×1.115 g=57.98 g EG

Moles of EG=57.98g ×62.07=0.9341 mol EG

Step 3. Calculate the mole fraction of water

Let water be component 1 and Ethylene Glycol be component 2.

$ntot =n1+n2$ = (2.363 + 0.9341) mol = 3.566 mol

$mole fraction of water =moles of water /total moles$ =2.362mol /3.566mol=0.6624

Step 4. Calculate the vapour pressure over the solution

$Pvap =p1=χ1p∘1$ = (0.6624 × 92) Torr = 61 Torr