Question 26 of 50intermediate🔧 ApplyNumerical3 marks

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Correct Answer

The vapour pressure of the mixture is 73.6 kPa.

Exercise: EXERCISES | Q: 1.16 | (Page 28)
For More Understanding

Explanation

This question applies Raoult's law for ideal solutions. Since heptane and octane form an ideal solution, the partial vapour pressure of each component equals the product of its mole fraction in the liquid phase and the vapour pressure of the pure component. The total vapour pressure is the sum of partial pressures of both components.

Solution Steps

  1. Step 1: Calculate molar masses Molar mass of heptane (C₇H₁₆) = (7×127 \times 12) + (16×116 \times 1) = 100 g mol⁻¹ Molar mass of octane (C₈H₁₈) = (8×128 \times 12) + (18×118 \times 1) = 114 g mol⁻¹

  2. Step 2: Calculate number of moles Moles of heptane (n₁) = 26.0 g / 100 g mol⁻¹ = 0.26 mol Moles of octane (n₂) = 35 g / 114 g mol⁻¹ = 0.307 mol Total moles = 0.26 + 0.307 = 0.567 mol

  3. Step 3: Calculate mole fractions Mole fraction of heptane (x₁) = 0.26 / 0.567 = 0.459 Mole fraction of octane (x₂) = 0.307 / 0.567 = 0.541

  4. Step 4: Apply Raoult's law Partial pressure of heptane (p₁) = x₁ × p₁° = 0.459×105.20.459 \times 105.2 kPa = 48.3 kPa Partial pressure of octane (p₂) = x₂ × p₂° = 0.541×46.80.541 \times 46.8 kPa = 25.3 kPa

  5. Step 5: Calculate total vapour pressure Total vapour pressure = p₁ + p₂ = 48.3 + 25.3 = 73.6 kPa

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