The vapor pressure of water at 25°C is 23.76 mm Hg. What would you calculate as the new vapor pressure of a solution made by adding 50.5 g of ethylene glycol (HOCH2CH2OH, antifreeze) to 50.5 g of water? You may assume that the vapor pressure of ethylene glycol at this temperature is negligible.

Answer 1

The first step to solve this equation is to calculate the moles of the solvent and solute.

The molar mass of H2O=18.016 g/mol
The molar mass of C2H6O2=62 g/mol

50.5g of C2H6O2*(1mol/62g)=0.815 mol C2H6O2

50.5 g of H2O*(1mol/18.016g)=2.803 mol H2O.

Now you need to find the mole fraction of the solvent. Since H20 has more moles in the solution, it acts as the solvent.

X=mol solvent/total mol
X=2.803/3.618=0.775

Now you can calculate the partial pressure of your solution
Psoln=XH2O*PH2O
Psoln=0.775*23.76=18.41 mmHg