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Solutions - Raoult's Law

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Raoult's Law states that for a solution of volatile liquids, the partial vapor pressure of each component of the solution is directly proportional to its mole fraction present in solution.

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For a solution containing a non-volatile solute, the vapor pressure of the solution is lower than that of the pure solvent. The relative lowering of vapor pressure is equal to the mole fraction of the solute: P0−PsP0=χsolute\frac{P^0 - P_s}{P^0} = \chi_{solute}.

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Ideal Solutions are those that obey Raoult's law over the entire range of concentration. For these solutions, enthalpy of mixing ΔmixH=0\Delta_{mix} H = 0 and volume change on mixing ΔmixV=0\Delta_{mix} V = 0.

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Non-ideal solutions show deviations from Raoult's law. Positive deviation occurs when the intermolecular forces between solute-solvent (A−BA-B) are weaker than those between solvent-solvent (A−AA-A) or solute-solute (B−BB-B).

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Negative deviation occurs when the intermolecular forces between solute-solvent (A−BA-B) are stronger than those between solvent-solvent (A−AA-A) or solute-solute (B−BB-B), often due to hydrogen bonding or chemical association.

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Azeotropes are binary mixtures having the same composition in liquid and vapor phase and boil at a constant temperature. Minimum boiling azeotropes show large positive deviation, while maximum boiling azeotropes show large negative deviation.

📐Formulae

PA=PA0χAP_A = P_A^0 \chi_A

Ptotal=PA+PB=PA0χA+PB0χBP_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B

Ptotal=PA0+(PB0−PA0)χBP_{total} = P_A^0 + (P_B^0 - P_A^0) \chi_B

P0−PsP0=n2n1+n2≈w2/M2w1/M1 (for dilute solutions)\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{w_2 / M_2}{w_1 / M_1} \text{ (for dilute solutions)}

yA=PAPtotal (Mole fraction of component A in vapor phase)y_A = \frac{P_A}{P_{total}} \text{ (Mole fraction of component A in vapor phase)}

💡Examples

Problem 1:

Vapor pressure of pure water at 298K298 K is 23.8mmHg23.8 mmHg. 50g50 g of urea (NH2CONH2NH_2CONH_2) is dissolved in 850g850 g of water. Calculate the vapor pressure of water for this solution and its relative lowering.

Solution:

  1. Moles of urea (n2n_2): 5060=0.833mol\frac{50}{60} = 0.833 mol.
  2. Moles of water (n1n_1): 85018=47.22mol\frac{850}{18} = 47.22 mol.
  3. Mole fraction of solute (urea) χ2\chi_2: 0.8330.833+47.22=0.0173\frac{0.833}{0.833 + 47.22} = 0.0173.
  4. Relative lowering P0−PsP0=χ2=0.0173\frac{P^0 - P_s}{P^0} = \chi_2 = 0.0173.
  5. Vapor pressure of solution Ps=P0(1−χ2)=23.8×(1−0.0173)=23.39mmHgP_s = P^0(1 - \chi_2) = 23.8 \times (1 - 0.0173) = 23.39 mmHg.

Explanation:

Applying the formula for relative lowering of vapor pressure for a non-volatile solute. The vapor pressure of the solution is slightly lower than that of pure water.

Problem 2:

At 350K350 K, the vapor pressure of pure liquids AA and BB are 450mmHg450 mmHg and 700mmHg700 mmHg respectively. Find the composition of the liquid mixture if the total vapor pressure is 600mmHg600 mmHg.

Solution:

Using the equation: Ptotal=PA0+(PB0−PA0)χBP_{total} = P_A^0 + (P_B^0 - P_A^0) \chi_B. 600=450+(700−450)χB600 = 450 + (700 - 450) \chi_B 150=250χB150 = 250 \chi_B χB=150250=0.6\chi_B = \frac{150}{250} = 0.6. Therefore, χA=1−0.6=0.4\chi_A = 1 - 0.6 = 0.4.

Explanation:

By substituting the known values of pure vapor pressures and total pressure into Raoult's law equation for binary volatile mixtures, we solve for the mole fraction of component BB and subsequently component AA.

Raoult's Law - Revision Notes & Key Formulas | ICSE Class 12 Chemistry