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Solutions

Question
CBSEENCH12005746

Benzene and toluene form iedal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene 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.

Solution
Solution
Given that
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



Mole of benzene (nA)
= Mass of bezeneMolar mass of bezene (C6H6)= 8078 g mol-1 = 1.025 mol.

Mass of Toluene (nB)
= Mass of tolueneMolar mass of toulene(C7H8)=1= 10092 g mol-1=1.087 mol

Mole fraction of benzene (xA)
= 1.025 mol(1.025 mol + 1.087mol) = 1.0252.113 = 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 

Use the formula of mole fraction using partial pressure

Mole fraction of benzene = PbenzenePbenzene + PToluene
  

Plug the values we get

Mole fraction of benzene = 24.645 /(24.645 +  16.48 )   
                                       
=  24.645/41.123  = 0.60