Question

A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be

• A. 0.178
• B. 0.278
• C. 0.378
• D. 0.478

Answer 1

The Correct Option is D

This problem involves calculating the mole fraction of pentane in the vapour phase of a solution that consists of a 1:4 mole ratio of pentane to hexane. We can solve this using Raoult's Law, which relates the partial pressure of a component in a mixture to its mole fraction in the liquid phase. 

1. \(n_{\text{pentane}} : n_{\text{hexane}} = 1:4\). Let's say we have 1 mole of pentane and 4 moles of hexane.
2. Calculate the mole fraction of each in the solution:
   • \(x_{\text{pentane}} = \frac{1}{1+4} = 0.2\)
   • \(x_{\text{hexane}} = \frac{4}{1+4} = 0.8\)
3. Using Raoult's Law, calculate the partial pressures:
   • \(P_{\text{pentane}} = x_{\text{pentane}} \times P^0_{\text{pentane}} = 0.2 \times 440 = 88 \text{ mm Hg}\)
   • \(P_{\text{hexane}} = x_{\text{hexane}} \times P^0_{\text{hexane}} = 0.8 \times 120 = 96 \text{ mm Hg}\)
4. Calculate the total pressure:
   • \(P_{\text{total}} = P_{\text{pentane}} + P_{\text{hexane}} = 88 + 96 = 184 \text{ mm Hg}\)
5. Now, find the mole fraction of pentane in the vapour phase (\(y_{\text{pentane}}\)):
   • \(y_{\text{pentane}} = \frac{P_{\text{pentane}}}{P_{\text{total}}} = \frac{88}{184} \approx 0.478\)

Thus, the mole fraction of pentane in the vapour phase is approximately \(0.478\), making the correct option

0.478

.