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Question

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


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Solution

Step 1: Given data

Ratio of Pentane to Hexane= 1:4

Partial pressure of Pentane= 440 mm Hg

Partial pressure of Hexane= 120 mm Hg

Step 2: Calculating total pressure

Mole fraction of Pentane is:

11+4=15

Mole fraction of Hexane is:

41+4=45

The total pressure of the solution is:

Ps=XpPp0+XhPh0Ps=15×440+45×120Ps=184nm

Step 3: Calculating mole fraction

The mole fraction of Pentane in the vapour phase is:

XpPpPs=4405184=88184=0.478

Therefore, The mole fraction of Pentane in the vapour phase would be 0.478.


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