Raoult's Law

Key Concepts

• The escaping tendency of a solvent is measured by its vapor pressure, which is dependent on temperature.
• Vapor pressure measures the concentration of solvent molecules in the gas phase.
• Assuming the solute is nonvolatile, the only particles in the gas phase are solvent molecules.
• In a solution, fewer solvent molecules are at the surface compared to the pure solvent, so a smaller proportion of solvent molecules will be in the gas phase and the vapor pressure for the solution is lower than that for the pure solvent.

  P_a < P_a^o
  where P_a = vapor pressure of the solution
  and     P_a^o = vapor pressure of pure solvent

• Raoult's Law states that for an ideal solution the partial vapor pressure of a component in solution is equal to the mole fraction of that component times its vapor pressure when pure:

  P_a = X_aP_a^o
  where P_a = vapor pressure of the solution
  and     P_a^o = vapor pressure of pure solvent
  and     X_a = mole fraction of the solvent

• The fractional vapor pressure lowering is equal to the mole fraction of the solute:

  X_b = (P_a^o - P_a) ÷ P_a^o
  where P_a = vapor pressure of the solution
  and     P_a^o = vapor pressure of pure solvent
  and     X_b = mole fraction of the solute

• Fractional vapor pressure lowering can be used to calculate molecular mass (formula weight) or molar mass of a solute.

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Example 1: Calculating the Vapor Pressure of a Solvent

Raoult's Law Problem:

1.00 g of nonvolatile sulfanilamide, C₆H₈O₂N₂S, is dissolved in 10.0 g of acetone, C₃H₆O.
The vapor pressure of pure acetone at the same temperature is 400 mm Hg.
Calculate the vapor pressure of the solution.

1. What is the question asking you to do?

   Calculate the vapor pressure of the solution.

   P_a = ? mm Hg

2. What information have you been given?

   mass(solute) = mass(C₆H₈O₂N₂S_(s)) = 1.00 g

   mass(solvent) = mass(C₃H₆O) = 10.0 g

   vapor pressure of pure solvent = P_a^o = 400 mm Hg

3. What is the relationship between what you know and what you need to find out?

   P_a = X_aP_a^o

   P_a^o = 400 mm Hg

   X_a = mole fraction of solvent
       = moles_(solvent) ÷ (moles_(solute) + moles_(solvent))

4. Perform the calculations:

   • Calculate moles of solute: n(C₆H₈O₂N₂S)
     n(C₆H₈O₂N₂S) = mass(C₆H₈O₂N₂S) ÷ molar mass(C₆H₈O₂N₂S)
         mass(C₆H₈O₂N₂S) = 1.00 g
         molar mass(C₆H₈O₂N₂S) = (6 x 12.01) + (8 x 1.008) + (2 x 16.00) + (2 x 14.01) + 32.06 = 172.204 g mol^-1

     moles(C₆H₈O₂N₂S) = 1.00 g ÷ 172.204 = 0.005807 mol

   • Calculate moles of solvent: n(C₃H₆O)
     n(C₃H₆O) = mass(C₃H₆O) ÷ molar mass(C₃H₆O)
         mass(C₃H₆O) = 10.0 g
         molar mass(C₃H₆O) = (3 x 12.01) + (6 x 1.008) + 16.00 = 58.078 g mol^-1

     moles(C₃H₆O) = 10.0 g ÷ 58.078 = 0.1722 mol

   • Calculate the mole fraction of the solvent: X_solvent
     X_solvent = moles_solvent ÷ (moles_solute + moles_solvent)
     X_a = moles(C₃H₆O) ÷ [moles(C₃H₆O) + moles(C₆H₈O₂N₂S)]
         = 0.1722 ÷ [0.1722 + 0.005807] = 0.9674
   • Calculate the vapor pressure: P_a
     P_a = X_aP_a^o
     P_a = 0.9674 x 400 mm Hg = 386.96 mm Hg = 387 mm Hg

5. Is your answer plausible?

   Addition of solute to solvent lowers the vapor pressure, that is,
   P_a < P_a^o
   where P_a is vapor pressure of solution and P_a^o is vapor pressure of pure solvent
   P_a^o = 400 mm Hg (given in question)
   P_a = 387 mm Hg (calculated)
   387 < 400 so our answer is plausible.

6. State your solution to the problem.

   Vapor pressure of solution is 387 mm Hg

Example 2: Calculating the Molecular Mass (Formula Weight) of a Solute

Raoult's Law Problem:

5.00 g of a nonvolatile compound was dissolved in 100 g of water at 30^oC.
The vapor pressure of the solution was measured and found to be 31.20 Torr.
The vapor pressure of pure water at 30^oC is 31.82 Torr.
Calculate the molar mass of the unknown solute.

Solution to Raoult's Law Problem:

1. What is the question asking you to do?

   Calculate the molar mass of the solute.

   molar mass_solute = molar mass (b) = ? g mol^-1

2. What information have you been given?

   mass(solute) = m(b) = 5.00 g

   mass(solvent) = m(a) = mass(H₂O) = 100 g (at 30°C)

   vapor pressure of solution = P_a = 31.20 Torr (at 30°C)

   vapor pressure of pure solvent = P_a^o = 31.82 Torr (at 30°C)

3. What is the relationship between what you know and what you need to find?

   (i) P_a = X_aP_a^o
       X_a = P_a ÷ P_a^o

   (ii) X_a = moles(a)/(moles(a) + moles(b))
       X_amoles(a) + X_amoles(b) = moles(a)
       X_amoles(b) = moles(a) - X_amoles(a)
       moles(b) = (moles(a) - X_amoles(a))/X_a

       Note: moles(a) = mass(H₂O) ÷ molar mass(H₂O)

   (iii) moles(b) = mass(b) ÷ molar mass (b)
       molar mass (b) = mass(b) ÷ moles (b)

4. Perform the calculations

   • Calculate mole fraction of solvent: X_a
     X_a = P_a ÷ P_a^o
     X_a = 31.20 ÷ 31.82 = 0.9805
   • Calculate moles of solute: moles(b)
     moles(b) = (moles(a) - X_amoles(a))/X_a
     moles(b) = ([100 g/(2 × 1.008 + 16.00) g mol^-1] - 0.9805×[100 g/(2 × 1.008 + 16.00) g mol^-1])/0.9805
     moles(b) = ([100/18.016] - (0.9805×[100/18.016])/0.9805
     moles(b) = (5.5066 - [0.9895 × 5.5066])/0.9805
     moles(b) = 0.1074/0.9805 = 0.1095 mol = 0.110 mol
   • Calculate molar mass solute: molar mass (b)
     molar mass (b) = mass(b) ÷ moles (b)
     molar mass (b) = 5.00 ÷ 0.110 = 45.45 g mol^-1

5. Is your answer plausible?

   Use an alternative method to calculate molar mass:

   • Calculate the mole fraction of the solute: X_b
     X_b = (P_a^o - P_a) ÷ P_a^o
     X_b = (31.82 - 31.2) ÷ 31.82 = 0.0195
   • Calculate moles of solvent: n(H₂O)
     n(H₂O) = mass ÷ molar mass
     n(H₂O) = 100 g ÷ (2 x 1.008 + 16.00) g mol^-1 = 5.5506 mol
   • Calculate moles of solute: X_solute
     X_solute = n_solute ÷ (n_solute + n_solvent)
     0.0195 = n_solute ÷ (n_solute + 5.5506)
     0.0195(n_solute + 5.5506) = n_solute
     0.0195(n_solute) + 0.1082 = n_solute
     0.1082 = n_solute - 0.0195(n_solute)
     0.1082 = (1 - 0.0195)(n_solute)
     0.1082 = 0.9805(n_solute)
     (n_solute) = 0.1082 ÷ 0.9805 = 0.110 mol
   • Calculate molar mass of solute: molar mass
     molar mass = mass ÷ moles
     molar mass = 5.00 g ÷ 0.110 mol = 45.45 g mol^-1

   
   This result is the same as we calculated above.

6. State your solution to the problem.

   Molar mass of solute is 45.45 g mol^-1