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Exercises · 1.38

Q.Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

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For an ideal binary solution, Raoult's law gives the partial pressures; the mole fraction in the vapour phase follows from Dalton's law. Converting masses to moles, then applying these laws yields ybenzene=0.598y_{\text{benzene}} = 0.598.

When two liquids form an ideal solution, each component's vapour pressure is simply proportional to its mole fraction in the liquid phase—that's Raoult's law. The vapour above the solution is a mixture of both components, and the composition of that vapour depends on how much each liquid contributes to the total pressure. The key insight is that the more volatile component (higher pure vapour pressure) will be enriched in the vapour relative to the liquid.

We need to find what fraction of the vapour is benzene when we mix specific masses of benzene and toluene.

1. Convert masses to moles

Benzene is C6H6\text{C}_6\text{H}_6 with molar mass Mbenzene=6(12)+6(1)=78 g/molM_{\text{benzene}} = 6(12) + 6(1) = 78 \text{ g/mol}.

Toluene is C7H8\text{C}_7\text{H}_8 with molar mass Mtoluene=7(12)+8(1)=92 g/molM_{\text{toluene}} = 7(12) + 8(1) = 92 \text{ g/mol}.

nbenzene=8078=1.026 moln_{\text{benzene}} = \frac{80}{78} = 1.026 \text{ mol}

ntoluene=10092=1.087 moln_{\text{toluene}} = \frac{100}{92} = 1.087 \text{ mol}

2. Calculate mole fractions in the liquid phase

Total moles in solution:

ntotal=1.026+1.087=2.113 moln_{\text{total}} = 1.026 + 1.087 = 2.113 \text{ mol}

Mole fraction of benzene in liquid:

xbenzene=1.0262.113=0.4856x_{\text{benzene}} = \frac{1.026}{2.113} = 0.4856

Mole fraction of toluene in liquid:

xtoluene=1−0.4856=0.5144x_{\text{toluene}} = 1 - 0.4856 = 0.5144

3. Apply Raoult's law to find partial pressures

For an ideal solution, the partial pressure of each component is:

pbenzene=xbenzene⋅pbenzene0=0.4856×50.71=24.63 mm Hgp_{\text{benzene}} = x_{\text{benzene}} \cdot p^0_{\text{benzene}} = 0.4856 \times 50.71 = 24.63 \text{ mm Hg}

ptoluene=xtoluene⋅ptoluene0=0.5144×32.06=16.49 mm Hgp_{\text{toluene}} = x_{\text{toluene}} \cdot p^0_{\text{toluene}} = 0.5144 \times 32.06 = 16.49 \text{ mm Hg}

4. Find total vapour pressure …

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