The vapor pressure of pure water at 70°C is 234 mm Hg. What is the vapor pressure depression of a solution of 110. g of CaCl2 in 215 g of water?
I can not figure out how to do this problem for the life of me! Any help would be wonderfull, thank you so much!!! I keep getting 17.9, but thats wrong and I don't know how to do it the right way.
This requires using Raoult's Law
P(soln) = X(solvent)*P(solvent)
where P(soln) is the vapor pressure of the resulting solution, X(solvent) is the mole fraction of solvent, and P(solvent) is the vapor pressure of the pure solvent. Using the mole fraction of the solute instead of the solvent is one of the most common errors students make with these problems.
First we need to find X(solvent) which is found by looking at the moles of solvent divided by the moles of solute + solvent.
215 g H2O (1 mol/18.01 g) = 11.9 mol H2O
110 g CaCl2 (1 mol CaCl2/111 g)(3 mol ions/1 mol CaCl2) = 2.97 moles of ions
Remember that for colligative properties, we have to determine the moles of particles (in this case, ions) not just the concentration of the solution
X(solvent) = 11.9 mol H2O/(11.9 + 2.97 mol)
X(solvent) = 0.800
Now to plug into Raoult's Law
P(soln) = 0.800 (234 mmHg)
P(soln) = 187 mmHg