# 2.38    Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $\inline 300 \; K$ are $\inline 50.71\; mm \; Hg$ and  $\inline 32.06\; mm \; Hg$  respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with  $\inline 100 \; g$  toluene.

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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