🌟 Raoult’s Law

 Certainly! Here's a more empathetic, student-friendly, and board-exam-oriented explanation of Raoult’s Law, perfect for Class 12 CBSE/RBSE students preparing for the Chemistry board exam:

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🌟 Raoult’s Law – Made Easy for Class 12 Boards

(Chapter 1: Solutions | NCERT Chemistry)

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👨‍🔬 Who Gave the Law?

In 1886, a French chemist named Francois Marte Raoult discovered something very useful:

In a mixture of volatile liquids, each liquid doesn't lose its identity.
It contributes to the overall vapour pressure according to how much of it is present in the solution.

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💡 Raoult’s Law – Class 12 Definition:

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

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🧪 Simple Formula (To Remember):

$$p_1 = x_1 \cdot p_1^0 \quad \text{and} \quad p_2 = x_2 \cdot p_2^0$$

Where:

• $p_1, p_2$ = partial vapour pressures of components 1 and 2

• $x_1, x_2$ = mole fractions of each in the liquid phase

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

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🧠 What You Need to Understand for Boards:

• Vapour pressure depends on composition (mole fraction).

• This law applies to ideal liquid–liquid solutions (like benzene + toluene).

• You should be able to use this law to calculate:

  • Partial pressures

  • Total vapour pressure of the solution

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🎯 Board Exam Tips:

✅ Always write the definition clearly.
✅ Include the formula with proper symbols.
✅ Give a simple example if asked.
✅ Explain that Raoult’s Law is the base for understanding colligative properties.

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🧪 Example for Practice:

If:

• $x_1 = 0.6$, $p_1^0 = 100$ mm Hg

• $x_2 = 0.4$, $p_2^0 = 80$ mm Hg

Then:

• $p_1 = 0.6 \times 100 = 60$ mm Hg

• $p_2 = 0.4 \times 80 = 32$ mm Hg

• Total vapour pressure = 60 + 32 = 92 mm Hg

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Would you like a summary chart, a one-mark question set, or a Raoult's Law derivation next?