Question

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 Kare 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. 

Answer 1

Now, no. of moles present in 80 g of benzene        =80/78mol=1.026

And, no. of moles present in 100 g of toluene     =100/92mol=1.087

 Mole fraction of benzene, xb                     =1.026/1.026+1.087=0486

And, mole fraction of toluene,                     x,=1-0.486=0514

 It is given that vapour pressure of pure benze       

Therefore, partial vapour pressure of benzene,   

=0.486x50.71

=24.645mmHg