At 100 degrees C, pure H2O has a vapor pressure of 760 torr. An aqueous solution has a concentration of 1.90 M MgCl2. The van't Hoff factor is 2.45. The density of the solution is 1.09 g/mL. Calculate the vapor pressure of the solution.

Answer 1

You need to find the mole fraction of H2O.
1 L of the solution has a mass of
1000 mL x 1.09 g/mL = 1090 grams.
There are 1.90 mols MgCl2 in that 1 L and that has a mass of
1.90 x molar mass MgCl2 = approx 170 but you need to do it more accurately. This is just an estimate.
Grams of H2O in that 1 L solution = 1090-170 = approx 900 g or 900/18 = approx 50 mols.
mols MgCl2 from above = ? and that x i = ? mols corrected for i.
mols H2O from above = ?
XH2O = mols H2O/total mols where total mols = mols H2O + corrected mols MgCl2.
Then
psolution = XH2O*PoH2O
Post your work if you get stuck.

Answer 2

VP(Soln) = VP(Solvent) - VP(Lowering Factor)

Using the following:
=> VP(Lowering Factor) = VP(LF)
=> VP(Solvent) = VP(Solv)
=> Mole Fraction Solute = X(Solute) = X(Solu)
=> van 't Hoff factor = (VHF)

VP(Soln) = VP(Solv) - VP(LF)
--VP(LF-ionic Soln) = VP(Solv)·X(Solu)·(VHF)
Therefore...
VP(Ionic Soln) = VP(Solv) - [(VP-Solv)(X-Solu)(VHF)]

Given:
VP(Solv) = 760Torr = 760 mmHg @100-deg.C
[MgCl2] = 1.90M
VHF = 2.45
Density Soln = 1.09 g/ml

VP(MgCl2 Soln) = VP(HOH)@100C - [VP(HOH)·X(MgCl2)·(VHF)]

X(MgCl2):
1.9M(MgCl2)= (1.9mol(MgCl2)/L Soln)
= 1.9mol(MgCl2)/1000-ml Soln)
= [1.9mol(MgCl2)/(1000-ml Soln)(1.09 g/ml Soln)]
= 1.9mol(MgCl2)/1090 gms Soln)

Grams MgCl2 = 1.90mol(95.22g/mol) = 180.92 gms MgCl2
Grams HOH = 1090 gms Soln - 180.92 gms MgCl2 = 909.08 gms HOH.
Moles HOH = (909.08 gms/18 gms/mol) = 50.5 moles HOH

X(MgCl2) = [(moles MgCl2)/(moles MgCl2) + moles(HOH)]
= [(1.90)moles/(1.9 + 50.2)moles] = 0.036

VP(ionic soln) = VP(solv) - [(VP-Solv)(X-Solu)(VHF)]
= (760-mmHg) - [(760-mmHg)(0.036)(2.45)]
= (760 - 75.24)mmHg = 684.76mmHg = 684.76Torr.