Asked by Rukky
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.
Answers
Answered by
Grace
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
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
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