Benzene and toluene form ideal 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.

Benzene and toluene form ideal 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

Given that

P^o_Benzene = 50.51 mm Hg  

P^o_Naphthalene = 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

Use formula 

Mole fraction of benzene, X_Benzene      = 1.0226 / (1.026 + 1.087) 

                                                          = 1.026/ 2.113  = 0.486

Similarly

Mole fraction of toluene, X _Toluene = 1  - X_Benzene  = 1  - 0.486 = 0.514

Use the formula if Henry law

P_A      = p^o_A  ×  X_A

Partial vapour pressure of benzene, P_Benzene = p^o_Benzene   ×  XB_enzene     

P_Benzene=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

 

Plug the values we get

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

                                       =  24.645/41.123  = 0.60