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.
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
To calculate the new vapor pressure of the solution, we can use Raoult's law. According to Raoult's law, the vapor pressure of a solution is proportional to the mole fraction of each component in the solution.
First, we need to calculate the mole fraction of water in the solution. To do this, we need to calculate the number of moles of water and ethylene glycol.
Given that the molecular weight of water (H2O) is 18.015 g/mol and the molecular weight of ethylene glycol (HOCH2CH2OH) is 62.07 g/mol, we can calculate the number of moles for each component.
For water:
Number of moles = mass of water / molecular weight of water
= 50.5 g / 18.015 g/mol
For ethylene glycol:
Number of moles = mass of ethylene glycol / molecular weight of ethylene glycol
= 50.5 g / 62.07 g/mol
Next, we need to calculate the mole fraction of water:
Mole fraction of water = moles of water / (moles of water + moles of ethylene glycol)
Then, we can use Raoult's law to calculate the new vapor pressure:
New vapor pressure = mole fraction of water * vapor pressure of pure water
Given that the vapor pressure of water at 25°C is 23.76 mm Hg, we can substitute the values and calculate the new vapor pressure of the solution.