2.38    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  toluene.

Answers (1)

Firstly, we will find the no. of moles of the given compounds.

No. of moles of benzene :

                                                =\frac{80}{78} = 1.026\ mol

and the no. of moles of toluene :

                                                 =\frac{100}{92} = 1.087\ mol.

Now we will find mol fraction of both:-

 Mole fraction of benzene :- 

                                                    =\frac{1.026}{1.026+1.087} = 0.486

and mole fraction of toluene : 

                                                    =1 - 0.486 = 0.514

Now, 

                        Ptotal    =   Pb  +  P

or                               = 50.71\times0.486\ + 32.06\times0.514\ = 24.65 +16.48

or                                = 41.13\ mm\ of\ Hg

Hence mole fraction of benzene in vapour phase is given by :

                                        = \frac{24.65}{41.13} = 0.60

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