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Raoult's law and vapour pressure: Class 12 practice

For an ideal solution of two volatile liquids, p_A = x_A p°_A and the total pressure is p_A + p_B. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Mole fraction

The idea behind raoult's law and vapour pressure

For an ideal solution of two volatile liquids, p_A = x_A p°_A and the total pressure is p_A + p_B. For a non-volatile solute, relative lowering of vapour pressure (p° − p)/p° = x_solute.

Step-by-step method

  1. Find mole fractions.
  2. Partial pressure = mole fraction × pure vapour pressure.
  3. Total = sum; or use relative lowering = x_solute.

Worked example 1 (Basic)

2 mol of A (p° = 580 mm Hg) is mixed with 5 mol of B (p° = 187 mm Hg). Find the total vapour pressure.

  1. x_A = 0.286, x_B = 0.714
  2. p = 165.71 + 133.57 = 299.29 mm Hg
  3. Answer: 299.29 mm Hg

Worked example 2 (Standard)

2 mol of A (p° = 110 mm Hg) is mixed with 5 mol of B (p° = 122 mm Hg). Find the total vapour pressure.

  1. x_A = 0.286, x_B = 0.714
  2. p = 31.43 + 87.14 = 118.57 mm Hg
  3. Answer: 118.57 mm Hg

Worked example 3 (Challenge)

A non-volatile solute (mole fraction 0.08) is dissolved in a solvent with p° = 86 mm Hg. Find the vapour pressure of the solution.

  1. (p° − p)/p° = 0.08
  2. p = 86(1 − 0.08) = 79.12 mm Hg
  3. Answer: 79.12 mm Hg

Common mistakes and misconceptions

  • Using mass fraction
  • Forgetting to add partial pressures

Practice questions

  1. (Basic) 4 mol of A (p° = 216 mm Hg) is mixed with 2 mol of B (p° = 162 mm Hg). Find the total vapour pressure.
  2. (Basic) 2 mol of A (p° = 478 mm Hg) is mixed with 5 mol of B (p° = 81 mm Hg). Find the total vapour pressure.
  3. (Standard) 3 mol of A (p° = 577 mm Hg) is mixed with 3 mol of B (p° = 122 mm Hg). Find the total vapour pressure.
  4. (Standard) 4 mol of A (p° = 270 mm Hg) is mixed with 2 mol of B (p° = 235 mm Hg). Find the total vapour pressure.
  5. (Challenge) A non-volatile solute (mole fraction 0.08) is dissolved in a solvent with p° = 53 mm Hg. Find the vapour pressure of the solution.
  6. (Challenge) A non-volatile solute (mole fraction 0.09) is dissolved in a solvent with p° = 100 mm Hg. Find the vapour pressure of the solution.

Answer key with full working

  • Q1: x_A = 0.667, x_B = 0.333 → p = 144 + 54 = 198 mm Hg → Answer: 198 mm Hg
  • Q2: x_A = 0.286, x_B = 0.714 → p = 136.57 + 57.86 = 194.43 mm Hg → Answer: 194.43 mm Hg
  • Q3: x_A = 0.5, x_B = 0.5 → p = 288.5 + 61 = 349.5 mm Hg → Answer: 349.5 mm Hg
  • Q4: x_A = 0.667, x_B = 0.333 → p = 180 + 78.33 = 258.33 mm Hg → Answer: 258.33 mm Hg
  • Q5: (p° − p)/p° = 0.08 → p = 53(1 − 0.08) = 48.76 mm Hg → Answer: 48.76 mm Hg
  • Q6: (p° − p)/p° = 0.09 → p = 100(1 − 0.09) = 91 mm Hg → Answer: 91 mm Hg

Syllabus note

This skill is listed under "Solutions" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is raoult's law and vapour pressure in?

"Solutions", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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