Vapour Pressure of Liquid–Liquid Solutions

 Here is a CBSE Class 12-friendly explanation of the full concept you've provided from Section 2.4.1: Vapour Pressure of Liquid–Liquid Solutions, including Raoult’s Law, Dalton’s Law, equations, graph interpretation, and key takeaways—ideal for notes, classroom explanation, or presentation.

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📘 2.4.1 Vapour Pressure of Liquid–Liquid Solutions

(Class 12 Chemistry – NCERT Chapter 1: Solutions)

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💧 What Are We Studying?

We are considering a binary solution made of two volatile liquids:

• Component 1 → less volatile

• Component 2 → more volatile

When both are kept in a closed container, they evaporate and reach a state of equilibrium between:

• Liquid phase and

• Vapour phase

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🔬 Raoult’s Law for Volatile Liquids

The partial vapour pressure of each component in a solution is directly proportional to its mole fraction in the liquid phase.

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📘 Mathematical Expressions:

For component 1:

$$p_1 = x_1 \cdot p_1^0 \tag{2.12}$$

For component 2:

$$p_2 = x_2 \cdot p_2^0 \tag{2.13}$$

Where:

• $p_1^0, p_2^0$: vapour pressures of pure components

• $x_1, x_2$: mole fractions in the liquid phase

• $p_1, p_2$: partial vapour pressures

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🧪 Total Vapour Pressure (Dalton’s Law):

$$p_{\text{total}} = p_1 + p_2 \tag{2.14}$$

Substitute from above:

$$p_{\text{total}} = x_1 p_1^0 + x_2 p_2^0 \tag{2.15}$$

Using $x_1 = 1 - x_2$:

$$p_{\text{total}} = p_1^0 + (p_2^0 - p_1^0) \cdot x_2 \tag{2.16}$$

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📈 Graphical Interpretation (Fig 2.3):

• A linear graph between:

  • $p_1$ vs $x_1$

  • $p_2$ vs $x_2$

  • $p_{\text{total}}$ vs $x_2$

The line:

• Starts from $p_1^0$ (when $x_2 = 0$)

• Ends at $p_2^0$ (when $x_2 = 1$)

📌 Assuming: $p_1^0 < p_2^0$ → component 1 is less volatile

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🔍 Key Conclusions from Raoult’s Law:

1. Total vapour pressure depends on the mole fraction of either component.

2. $p_{\text{total}}$ varies linearly with composition in an ideal solution.

3. Increasing mole fraction of more volatile component → increases total vapour pressure.

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🌫️ Vapour Phase Composition (Dalton’s Law again):

Let:

• $y_1, y_2$: mole fractions of components 1 and 2 in the vapour phase
  Then:

$$p_1 = y_1 \cdot p_{\text{total}} \tag{2.17}$$ $$p_2 = y_2 \cdot p_{\text{total}} \tag{2.18}$$

So in general:

$$p_i = y_i \cdot p_{\text{total}} \tag{2.19}$$

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🧠 Important Concept:

The vapour phase is richer in the more volatile component
because it contributes more to the total vapour pressure.

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📝 CBSE Board Exam Tips:

• State and apply Raoult’s Law

• Derive total vapour pressure using mole fractions

• Be able to interpret the graph (Fig. 2.3)

• Relate vapour phase composition to partial pressure

• Use the formulas with correct units

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