Ncert Exercises & Solutions

2.16 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?

Vapour pressure of heptane $(p_{1}^{0}) = 105.2 kPa$

Vapour pressure of octane $(p_{2}^{0}) = 46.8 kPa$

We know that,

Molar mass of heptane (C₇H₁₆) = 7 × 12 + 16 × 1 = 100 g mol⁻¹

$∴ Number of moles of heptane = \frac{26}{100} mol = 0.26 mol$

Molar mass of octane (C₈H₁₈) = 8 × 12 + 18 × 1 = 114 g mol⁻¹

$∴ Number of moles of octane = \frac{35}{114} mol = 0.31 mol$

$Mole fraction of heptane, x_{1} = \frac{0.26}{0.26 + 0.31} = 0.456$

$And, mole fraction of octane, x_{2} = 1 - 0.456 = 0.544$

$Now, partial pressure of heptane, p_{1} = x_{1} p_{1}^{0} = 0.456 \times 105.2 = 47.97 kPa$

$Partial pressure of octane, p_{2} = x_{2} p_{2}^{0} = 0.544 \times 46.8 = 25.46 kPa$

Hence, vapour pressure of solution, p_total = p₁ + p₂ = 47.97 + 25.46 = 73.43 kPa

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