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`        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.`
one year ago

```							Dear Rajesh  PoBenzene = 50.51 mm Hg PoNaphthalene = 50.51 mm HgMass of Benzene = 80 gMass of Toluene = 100 gMolar mass of benzene(C6H6) = 6 × 12 + 6 × 1 = 78 g mol - 1Molar mass of toluene(C6H5CH3) = 6 × 12 + 5 × 1 + 12 + 3 × 1 = 92 g mol – 1Use the formulaNumber of moles of benzene = 80 / 78 = 1.026 molNumber of mole of toluene = 100 / 9 2 = 1.087 molMole fraction of benzene, XBenzene = 1.0226 / (1.026 + 1.087) = 1.026/ 2.113 = 0.486SimilarlyMole fraction of toluene, X Toluene = 1 - XBenzene = 1 - 0.486 = 0.514Use the formula if Henry lawPA = poA × XAPartial vapour pressure of benzene, PBenzene = poBenzene × XBenzene PBenzene=0.487 × 50.71 = 24.645 mm HgSimilarlyPartial pressure of Toluene, P toluene = 0.514 × 32.06 = 16.48 mmHgMole fraction of benzene = 24.645 /(24.645 + 16.48 )= 24.645/41.123 = 0.60 RegardsArun (askIITians forum expert)
```
one year ago
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