Raoult's law

 

Describe Raoult’s law when solute is non-volatile.

Describe Raoult’s law when components are volatile.

Raoult’s law:

When solutes are dissolved into the solvents, then their vapour pressures are lowered. This lowering of vapour pressure is due to the presence of particles of the solute on the surface of solutions. We obtain quantitative relationship between vapour pressures of the solution and the quantities of the solutes and the solvents through Raoult’s law.

When solute is non-volatile, non-electrolyte:

Let us consider,

Vapour pressure of pure solvent is P^O

Vapour pressure of solution is P

Mole fraction of solvent is X₁

Mole fraction of solute is X₂

According to Raoult’s law “the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.”

             P ∝ X₁

             P = P^OX₁                                                              eq.. (1)

Since X₁ +X₂= 1 (sum of the mole fractions of components of a solution is unity)                                                           eq.. (2)

Rearranging X₁ = 1 – X₂

Putting the value of X₁ in eq (1)

       P = P^O(1 – X₂)                                       eq.. (3)

       P = P^O – P^OX2

       P – P^O = P^OX₂

       ∆P = P^OX₂

           ∆P ∝ X₂

 ∆P is called lowering of vapour pressure.

         “ the lowering of vapour pressure is directly proportional to the mole fraction of the solute .”

      ∆P/P = X₂                                 eq.. (4)

∆P/P  is called relative lowering of vapour pressure.

 “the relative lowering of vapour pressure  is equal to the mole fraction of the solute.”

Importance of relative lowering of vapour pressue over lowering of vapour pressure:

(a) Independent of temperature(mole fraction is independent of temperature)

(b) Is a constant quantity, when equimolar proportions of different solutes are dissolved in same mass of the same solvent, e.g. in a solution with 3 moles of water and 3 moles of ethanol the mole fraction of each component is 3/6= 0.5 or 50%.

Graphical representation:

when solute is non-volatile we get a straight line

 

Raoult’s law when both components are volatile:

In a mixture of volatile liquids A and B

       P_A = P_A^OX_A

       P_B = P_B^OX_B

            P_total =  P_A + P_B

        P_total = P_A^OX_A + P_B^OX_B

            Where P_A^O and P_B^O are vapour pressures of pure liquid A and pure liquid B and P_A and P_B are partial pressures of A and B in solution. Thus, it is a special case of Dalton’s law of partial pressure of gases applied to volatile liquids.

          X_A + X_B = 1

So, X_B = 1 - X_A , P_total = P_A^OX_A + P_B^O(1 - X_A )

P_total = P_A^OX_A + P_B^O - P_B^OX_A = P_B^O + (P_A^O – P_B^O)X_A  eq.. (5)

 

Also, P_total = P_A^O + (P_B^O – P_A^O)X_B                                eq.. (6)

Mole fraction of component  A in vapour phase

                     X^’_A= P_A^OX_A/P_total

                     X^’_B =  P_B^OX_B/P_total

Graphical representation:

 

when both components are volatile

equations (5) and (6) are straight line equations. X_A is independent variable while P_total  is dependent. According to the graphical representation component A is more volatile as compared to B. So the vapour pressure of A(P_A^O) is greater than B(P_B^O ). We get a straight line if the solutions are ideal.

 

Important MCQs:

1.   A non-volatile solute(A) is dissolved in a volatile solvent B. The vapour pressure of the resultant solution is P_s and of the pure solvent is  P_B^O. If X_B is the mole fraction of the solvent in solution, which of the following is correct?

(a) P_s = X_AP_B^O                                   (b)P_B^O =X_BP_s

(c)P_s = X_BP_B^O                                    (d)P_s^O = X_APs

Ans. (c)

2.   The vapour pressure of benzene at a certain temperature is 640mm. A non-volatile and non-electrolyte solid weighing 2.175g is added to 39.08g of benzene. If the vapour pressure of the solution is 600mm, what is the molecular mass of the substance?

(a)49.50                               (b)59.60

(c)65.14                               (d)79.80

Ans. (c)

Solution:

We know that       ∆P/P = X₂          eq..(A)

X₂ is the mole fraction of solute which is equal to n₂/n₁+n₂

                                       X₂ =n₂/n₁+n₂

n₂= number of moles of solute

n₁= number of moles of solvent here benzene

mass of solute = 2.175g

mass of solvent/benzene = 39.08g

we know n₂= mass/molar mass = 2.175g/m(unknown)

               n₁= mass/molar mass= 39.08/78= 0.5010

 here P = vapour pressure of solution= 640mm

         P^O = vapour pressure of pure solvent= 600mm

        ∆P = P - P^O = 640mm-600mm= 40mm

Putting all the values in eq (A)

∆P/P = n₂/n₁+n₂ = (2.175g/m)/0.5010+2.175g/m

40mm/640mm =(2.175g/m)/0.5010

0.0625(0.5010+ 2.175g/m) =2.175g/m

0.0313 +0.1359/m= 2.175g/m

0.0313=2.175g/m -0.1359/m

0.0313=(2.175-0.1359)/m= 2.0391/m

m= 2.0391/0.0313= 65.14g

3.   The vapour pressure of water at 300K is 0.4atm in a closed container. If the volume of container is doubled its vapour pressure at 300K will be

(a)0.08atm                  (b)0.2atm

(c)0.6atm                    (d)0.4 atm

Ans. (d) vapour pressure depends upon temperature not volume of container