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Solutions - Vapour Pressure of Liquid Solutions (Raoult's Law)

Grade 12CBSEChemistry

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

🔑Concepts

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Vapour Pressure: The pressure exerted by the vapours of a liquid in equilibrium with the liquid at a given temperature is called its vapour pressure. It depends on the nature of the liquid and the temperature.

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Raoult's Law for Volatile Liquids: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in solution. For a binary solution, p1=p10x1p_1 = p_1^0 x_1 and p2=p20x2p_2 = p_2^0 x_2.

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Raoult's Law for Non-volatile Solutes: When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lowered. The relative lowering of vapour pressure is equal to the mole fraction of the solute: p10−pp10=x2\frac{p_1^0 - p}{p_1^0} = x_2.

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Ideal Solutions: Solutions that obey Raoult's law over the entire range of concentration. For these, ΔHmixing=0\Delta H_{mixing} = 0 and ΔVmixing=0\Delta V_{mixing} = 0. Examples: nn-hexane and nn-heptane, bromoethane and chloroethane.

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Non-ideal Solutions: Solutions that do not obey Raoult's law. They show either Positive Deviation (intermolecular forces A−BA-B are weaker than A−AA-A and B−BB-B) or Negative Deviation (intermolecular forces A−BA-B are stronger than A−AA-A and B−BB-B).

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Azeotropes: Constant boiling mixtures that have the same composition in liquid and vapour phases. Minimum boiling azeotropes show large positive deviation (e.g., 95%95\% ethanol), while maximum boiling azeotropes show large negative deviation (e.g., 68% HNO368\% \text{ } HNO_3).

📐Formulae

p1=p10x1p_1 = p_1^0 x_1

Ptotal=p1+p2=p10x1+p20x2P_{total} = p_1 + p_2 = p_1^0 x_1 + p_2^0 x_2

Ptotal=p10+(p20−p10)x2P_{total} = p_1^0 + (p_2^0 - p_1^0)x_2

yi=piPtotal (Mole fraction in vapour phase)y_i = \frac{p_i}{P_{total}} \text{ (Mole fraction in vapour phase)}

p10−psp10=x2=n2n1+n2\frac{p_1^0 - p_s}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2}

💡Examples

Problem 1:

Vapour pressure of chloroform (CHCl3CHCl_3) and dichloromethane (CH2Cl2CH_2Cl_2) at 298 K298 \text{ K} are 200 mm Hg200 \text{ mm Hg} and 415 mm Hg415 \text{ mm Hg} respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g25.5 \text{ g} of CHCl3CHCl_3 and 40 g40 \text{ g} of CH2Cl2CH_2Cl_2 at 298 K298 \text{ K}.

Solution:

Molar mass of CH2Cl2=12+2(1)+2(35.5)=85 g/molCH_2Cl_2 = 12 + 2(1) + 2(35.5) = 85 \text{ g/mol}. Moles of CH2Cl2(n1)=4085=0.47 molCH_2Cl_2 (n_1) = \frac{40}{85} = 0.47 \text{ mol}. Molar mass of CHCl3=12+1+3(35.5)=119.5 g/molCHCl_3 = 12 + 1 + 3(35.5) = 119.5 \text{ g/mol}. Moles of CHCl3(n2)=25.5119.5=0.213 molCHCl_3 (n_2) = \frac{25.5}{119.5} = 0.213 \text{ mol}. Total moles =0.47+0.213=0.683 mol= 0.47 + 0.213 = 0.683 \text{ mol}. xCH2Cl2=0.470.683=0.688x_{CH_2Cl_2} = \frac{0.47}{0.683} = 0.688. xCHCl3=1−0.688=0.312x_{CHCl_3} = 1 - 0.688 = 0.312. Ptotal=pCH2Cl20xCH2Cl2+pCHCl30xCHCl3P_{total} = p^0_{CH_2Cl_2}x_{CH_2Cl_2} + p^0_{CHCl_3}x_{CHCl_3} Ptotal=(415×0.688)+(200×0.312)=285.5+62.4=347.9 mm HgP_{total} = (415 \times 0.688) + (200 \times 0.312) = 285.5 + 62.4 = 347.9 \text{ mm Hg}.

Explanation:

We first calculate the number of moles for each component to find their mole fractions in the liquid phase. Then, Raoult's Law is applied to find the partial pressures of each component, the sum of which gives the total vapour pressure.

Problem 2:

The vapour pressure of pure water at 298 K298 \text{ K} is 23.8 mm Hg23.8 \text{ mm Hg}. 50 g50 \text{ g} of urea (NH2CONH2NH_2CONH_2) is dissolved in 850 g850 \text{ g} of water. Calculate the vapour pressure of water for this solution.

Solution:

Moles of urea (n2)=5060=0.833 mol(n_2) = \frac{50}{60} = 0.833 \text{ mol}. Moles of water (n1)=85018=47.22 mol(n_1) = \frac{850}{18} = 47.22 \text{ mol}. Mole fraction of water (x1)=47.2247.22+0.833=47.2248.053≈0.983(x_1) = \frac{47.22}{47.22 + 0.833} = \frac{47.22}{48.053} \approx 0.983. Vapour pressure of solution (ps)=p10x1=23.8×0.983=23.4 mm Hg(p_s) = p_1^0 x_1 = 23.8 \times 0.983 = 23.4 \text{ mm Hg}.

Explanation:

Urea is a non-volatile solute. According to Raoult's Law, the vapour pressure of the solution is determined by the mole fraction of the solvent (water) multiplied by the vapour pressure of pure solvent.

Vapour Pressure of Liquid Solutions (Raoult's Law) Revision - Class 12 Chemistry CBSE