Benzene and toluene form nearly ideal solution. At `298 K`, the vapour pressure of pure benzene is 150 torr and of pure toluence is 50 torr. Calculate the vapour pressure of the solution, containing equal weights of two substances at this temperature?

Answer

104.1 torr

Explanation

Let the weight of benzene and toluene in the solution = w gram .
Moles of benzene ` = (w)/(78)` (Molar mass = 78)
Moles of toluene ` = (w)/(92) ` (Molar mass = 92)
Mole fraction of benzene ,`x_(b) = ((w)/(78))/((w)/(78) + (w)/(92)) = 0.541`
Mole fraction of toluene , ` x_(t) = 1 - 0.541 = 0.459`
Partial vapour pressure of benzene ,
` p_(b) = P_(b)^(@) xx x_(b) = 150 xx 0.541 = 81.15` torr .
Partial vapour pressure of toluene ` p_(t) = p_(t)^(@) xx x_(t) = 50 xx 0.459`
` = 22.95` torr.
Total vapour pressure of solution ` = 81.15+22.95`
` = 104.1` torr.