Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile solute dissolved in 78 gram benzene, benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. [Molar mass of benzene is 78 g/`mol^(-1)`]

Explanation

(a) `M_(2) = (W_(2) xx M_(1) xx P_(1)^(0))/((P_(1)^(0) - P_(a)) xx W_(1))`
`M_(2) = (2 xx 78 xx 200)/((200 - 195) xx 78)`
`M_(2) = 80 g//mol`
(b) Binary liquid mixtures having the same composition in liquid and vapour phase and boil at a constant temperature.
or
Constant boiling point liquid mixture
[Any suitable example for solution minimum boiling azeotrope.]
benzene and acetone, n-Hexane and ethanol, water and ethanol, acetone and `CS_(2), C Cl_(4)` and `CHCl_(3), C Cl_(4)` and Toluene, Acetone and ethanol.
(a) Given * Vapour pressure of pure benzene `(P_(0))` = 200 mm of Hg.
* Mass of non-volatile solute `(W_(B) = 2g`
* Mass of benzene as solvent `(W_(A)) = 78 g`
* Molar mass of benzene `(M_(A)) = 78 g mol^(-1)`.
* Vapour pressure of solution (P) = 195 mm of Hg
`because` No. of moles (n) `= ("W(mass)")/("(M(Molar mass)")`
`:.` Moles of benzene (n) `= (W)/(M) = (78)/(78) = 1`
and Moles of non-volatile solute `(n_(B)) = (W_(B))/(M_(B)) = (2)/(M_(B))`
To find : Value of `M_(B)`.
also
`(Delta p)/(p^(0)) = (n_(B))/(n_(A))`
`(p^(0) - p)/(p^(0)) = (n_(B))/(n_(A))`
`(Delta p)/(p^(0)) = (200 - 195)/(200) = ((2)/(M_(B)))/((1)/(1))`
`:. (5)/(200) = (2)/(M_(B))`
or `M_(B) = (2 xx 200)/(5) = 80`
Molar mass of solute `= 80 g mol^(-1)`.
(b) Azotropes : it is a mixture of two liquids which has a constant boiling point and composition through out distillation. It is of two types :
The minimum boiling azeotrope and the maximum boiling azeotrope.
A solution that shows (+) ve deviation from Raoult.s law is called minimum boiling zeotrope while a solution that shows (-) ve deviation from Raoult.s law is called maximum boiling azeotrope.
Mixture of ethanol and water, in which ethanol is approximately 95% by volume is an example of minimum boiling azeotrope.