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Grade: 11
        
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.
10 months ago

Answers : (1)

Arun
14776 Points
							
Dear Rajesh
 
 
PoBenzene = 50.51 mm Hg 
PoNaphthalene = 50.51 mm Hg
Mass of Benzene = 80 g
Mass of Toluene = 100 g
Molar mass of benzene(C6H6) = 6 × 12 + 6 × 1 = 78 g mol - 1
Molar mass of toluene(C6H5CH3) = 6 × 12 + 5 × 1 + 12 + 3 × 1 = 92 g mol – 1
Use the formula

Number of moles of benzene = 80 / 78 = 1.026 mol
Number of mole of toluene = 100 / 9 2 = 1.087 mol
Mole fraction of benzene, XBenzene = 1.0226 / (1.026 + 1.087) 
= 1.026/ 2.113 = 0.486
Similarly
Mole fraction of toluene, X Toluene = 1 - XBenzene = 1 - 0.486 = 0.514
Use the formula if Henry law
PA = poA × XA
Partial vapour pressure of benzene, PBenzene = poBenzene × XBenzene 
PBenzene=0.487 × 50.71 = 24.645 mm Hg
Similarly
Partial pressure of Toluene, P toluene = 0.514 × 32.06 = 16.48 mmHg

Mole fraction of benzene = 24.645 /(24.645 + 16.48 )
= 24.645/41.123 = 0.60
 
Regards
Arun (askIITians forum expert)
10 months ago
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