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Question:
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.

Answer:

Molar mass of benzene(C₆H₆) = 6 X 12 + 6 X 1 = 78 g/mol

Molar mass of toluene = 7 x 12 + 8 x 1 = 92 g/mol

Now no of moles in 80g of benezen = 80 / 78 = 1.026 mol

No of moles in 100g of toluene = 100 / 92 = 1.087 mol

∴Mole fraction of benzene x_b = 1.026 / 1.026 + 1.087 = 0.486

And Mole fraction of toluene,x_t = 1 - 0.486 = 0.514

We have given that

Vapor pressure of pure benzene p_b° = 50.71 mm Hg

And, vapour pressure of pure toluene, p_t° = 32.06 mm Hg

Therefore partial Vapor pressure of benzene, p_b = p_b X x_b

= 50.71 x 0.486

= 24.65 mm Hg

And partial Vapor pressure of toluene, p_t = p_t X x_t

P_t = p°_t X x_t = 32.06 x 0.514

= 16.48

Total vapour pressure = 24.65 + 16.48 = 41.13 mm Hg

Mole fraction of benzene in vapour phase = 24.65 / 41.13 = 0.60

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Question:
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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Question: 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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Question:
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.