Solution and Colligative properties

Vapour pressure of liquid solution:- 
      when a liquid solution is kept in a closed when a container, some of the molecules of liquid from its surface starts evaporating & start accumulating in the space above the surface of the liquid. due to lesser force of attraction and more kinetic energy present in surface molecule.
     Molecules of liquid in the vapour phase more randomly & strike the liquid surface & get condensed. The process keeps on until equilibrium between evaporation & condensation reached.            
      The pressure exerted by molecules of vapour phase present in the vacant space of the container over the liquid surface is called vapour pressure.
Factors on which vapour pressure of a liquid depends : - 
( a). Nature of the liquid - Each liquid has a characteristics vapour pressure because each liquid has different magnitude of intermolecular forces .
(b). Temperature - The vapour pressure of a liquid increase with increase in temperature. due to increase in kinetic energy of molecules present in liquid(with increase in temperature).
 (c).Boiling point:-vapour pressure is inverse  to boiling point.
(d). Purity of liquid : pure liquid has always higher vapour pressure compare to its solution .
Vapour Pressure of liquid - liquid solutions:- 
                The French chemist, Francois Mate Raoult (1886)gave a quantitative relationship between partial vapour pressure of each component of the solution & its mole fraction.
It states that the vapour pressure of solution is directly proportional to mole faction of a solute added to the solution
Raoult law for volatile liquids: -It states that The partial pressure of each volatile component of the solution is directly proportional to its mole fraction present in solution.
let us consider a binary solution of two volatile liquids denoted as component-1 & 2. Let the total vapour pressure be Ptotal at equilibrium state, p₁ & p₂ be the partial vapour pressure of two components 1 & 2 respectively.
Thus, by applying RaouIt's law for component - 1
               p₁ α  𝝌₁               . . . . . . . . .(i)
                    P₁ = p₁⁰ 𝝌_{1           . . . . . . . . . . . .(ii)}   where p₁⁰ is the vapour pressure of  pure component -1  

similarly for component -2
                   p₂ α  𝝌₂. . . . . . . . .(iii)
                  P₂ = p₂⁰ 𝝌_{2   . . .  . . . .(iv)}       where p₁⁰ is the vapour pressure of  pure component -2
According to Dalton's law of partial pressures-
                                                                            the total pressure (p _total)over the solution phase will be the sum of the partial pressure of the components of the solution.

   p_{total =}  p_{1 +} p_{2  . . . . . . . . . . . . . . . . (v)  
           by} substituting the values of p_{1 &} p_{2 we get} 

    p_total=p₁⁰ 𝝌_1 +  p₂⁰ 𝝌_{2     . . . . . . . . .(vi)                                                                             
             = (1-}𝝌₂₎ p₁⁰₊  p₂⁰ 𝝌_{2  . . . . . . . (vii) Since}  𝝌_1 +  𝝌_{2 = 1
            =}   p₁⁰₊₍ p₂⁰ _- p₁⁰ )𝝌_{2 . . . . . . .(viii)}
for-composition of vapour phase   in equilibrium: 
                 Let  y₁ &  y₂mole fraction and p_1,p₂ partial pressure of the component 1 & 2 respectively in vapour phase.
P₁ =  y₁ p_{total       . . . . . (ix)}          

P₂ =  y₂ p_{total     . . . . . .(x)}

or   y₁₌ p₁/p_total                 y₂₌  p₂ / p_totat 
    y₁= p₁ 𝝌₁⟋p⁰₁ 𝝌_1 +p₂⁰𝝌₂    . . . . . . . ( xi)    
     y₂= p₂ 𝝌₂/p⁰₁ 𝝌_1 +p₂⁰𝝌_{2  . . . . . . . . . xii
eq. ( xi) & (xii) may be used to find out the composition of  in vapour phase.}
Raoults Law for solutions containing solids in liquids: 
 Only liquids are volatile components in solution. So it can evaporate as vapour. Thus thus vapour pressure is due to volatile liquid. 
Raoult's law may state that the partial vapour pressure of volatile component in solution is directly proportional to its mole fraction. 
In mathematics 
   p_vol  α  𝝌_vol
   P_vol = p_vol⁰ 𝝌_{vol 
Solid non volatile does not contribute in vapour pressure, since it does not form vapour.}  

Ideal and non ideal solutions:-

              Ideal solution                                                
 i)  Obey Raoults law P₁ = p₁⁰ 𝝌₁   P₂ = p₂⁰ 𝝌₂  

 ii)  ΔH (mix) = 0            ΔV (mix) = 0 

              non ideal solution     

 i) don't obey raoults law   P₁ # p₁⁰ 𝝌₁    P₂ # p₂⁰ 𝝌₂ 

ii)   ΔH (mix) ≠ 0                    ΔV (mix) ≠ 0

                       

ideal

non ideal solution

Independent For Example, consider two liquids A and B, and mix them. The formed  solution will experience several intermolecular forces of attractions inside it, which will be: A – A intermolecular forces of attraction B – B intermolecular forces of attraction A – B intermolecular forces of attraction The solution is said to be an ideal solution, only  when the intermolecular forces of attraction  between A – A, B – B and A – B are nearly equal. Examples of Ideal Solutions n-hexane and n-heptane Bromoethane and Chloroethane Benzene and Toluene Chlorobenzene and Bromobenzene CCl4 and SiCl4

Interacts with others.                           The solute-solute and solvent-solvent interaction is different from that of solute-solvent interaction Non-ideal solutions are of two types: (a) Non-ideal solutions showing positive deviations:  Positive deviation occurs when total vapour pressure for any mole fraction is more than what is expected according to Raoult’s law. This happens when the new interactions are weaker than the interaction in the pure component (A – B < A – A or B – B interactions). It forms minimum boiling azeotropes, for example, C2H5OH + cyclohexane. The Bonding present in pure C2H5OH is cut off on adding cyclohexane. For such solution, ΔV and ΔH are positive. Examples: Acetone + carbon disulphide, Acetone + benzene Carbon tetrachloride + chloroform or Toluene Methyl alcohol + water Acetone + C2H5OH Non-ideal solutions showing negative deviations:  Negative deviation occurs when total vapour pressure for any mole fraction is less than what is expected according to Raoult’s law. This happens when the new interactions are stronger than the interaction in the pure component (A – B > A – A or B – B interactions). It forms maximum boiling azeotrope, for example, CHCl3+ CH3COCH3.For such solutions, ΔV and ΔH are negative. Examples: Chloroform + benzene or diethyl ether Acetone + aniline Nitric acid (HNO3) + water Acetic acid + pyridine