Question 27 of 50intermediate🔧 ApplyNumerical2 marks

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Correct Answer

12.08 kPa (approximately)

Exercise: EXERCISES | Q: 1.17 | (Page 28)
For More Understanding

Explanation

This problem applies Raoult's law for solutions containing non-volatile solutes. The vapour pressure of the solution is directly proportional to the mole fraction of the solvent. A 1 molal solution means 1 mole of solute dissolved in 1 kg (1000 g) of solvent (water).

Solution Steps

  1. Step 1: Given - Vapour pressure of pure water (p₁⁰) = 12.3 kPa, Molality = 1 molal

  2. Step 2: For 1 molal solution: Moles of solute (n₂) = 1 mol, Mass of water = 1000 g

  3. Step 3: Moles of water (n₁) = 1000/18 = 55.56 mol

  4. Step 4: Mole fraction of solvent (x₁) = n₁/(n₁ + n₂) = 55.56/(55.56 + 1) = 55.56/56.56 = 0.9823

  5. Step 5: Applying Raoult's law: p₁ = x₁ × p₁⁰ = 0.9823×12.30.9823 \times 12.3 = 12.08 kPa

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