Question

benzene and toulene form ideal solution the vapour pressure of pure benzene and toulene at 300 k are 50.71 mm and 32.06 mm hg. Calculate the mole fraction of benzene in vapour phase if 80 gm of benzene is mined with 100 gm toulene
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Answer 1

Answer:

.60

Explanation:

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 xb = 1.026 / 1.026 + 1.087 = 0.486

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

pb° = 50.71 mm Hg

pt° = 32.06 mm Hg

Therefore partial Vapor pressure of benzene, pb = pb X xb

= 50.71 x 0.486

= 24.65 mm Hg

And partial Vapor pressure of toluene, pt = pt X xt

Pt = p°t X xt = 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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