Chemistry - Liquid Solution Question with Solution | TestHub

The pressure exerted by the vapours above the liquid surface in equllibrium with the liquid at a given temperature is called vapour pressure.  
    
    When a non-volatile solute is dissolved in a solvent, its vapour pressure decreases. This decrease in vapour pressure is known as “Lowering of vapour pressure”. The ratio of the  Lowering of vapour pressure to the vapour pressure of the pure solvent is called Relative lowering of vapour pressure. It is a Colligative Property.
    Statement:  Raoult's law  states that “at a given temperature, for a solution of volatile liquids the partial vapour pressure  of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction”.${\mathrm{P}}_{\mathrm{solution}} \ne {\mathrm{P}}_{{}^0\mathrm{solvent}} \times {\mathrm{X}}_{\mathrm{solvent}}$
    For an ideal solution, ${\mathrm{P}}_{\mathrm{S}} = {\mathrm{P}}_{\mathrm{A}} + {\mathrm{P}}_{\mathrm{B}} = {{\mathrm{P}}_{\mathrm{A}}}^0 \times {\mathrm{X}}_{\mathrm{A}} + {{\mathrm{P}}_{\mathrm{B}}}^0 \times {\mathrm{X}}_{\mathrm{B}}$
    
    Definition: An ideal solution is one which obeys  Raoult's law  over the entire range of concentration.

 A non-ideal solution does not obey Raoult's law  as $P_{solution} \ne P_{{}^{0}solvent} \times X_{solvent}$  The vapour pressure of such a solution is either higher or lower than the product. When it is higher, the solution exhibits a positive deviation. When it is  lower, it exhibits a negative deviation from Raoult’s law.

 A positive deviation from Raoult’s Law arises when the forces between the particles in the mixture
 are weaker than the mean of the forces between the particles in the pure solvent. There is some
hydrogen bonding in pure ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$. However, when ${\mathrm{CS}}_2$ is added to  ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$, the molecules of ${\mathrm{CS}}_2$ tend to occupy the space  between ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$  molecules. Consequently some hydrogen bonds in ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$ molecules break and the attractive forces in ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$ molecules get weakened.
The escaping tendency of ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CO}$  and ${\mathrm{CS}}_2$   molecules from the solution will increase.  Because of this, effective
 number of particles will decrease in the solution. Hence vapour pressure of the solution will be
 greater than the vapour pressure as expected from Raoult’s law.  So this will show positive deviation from Raoult’s law.