Raoult's Law | Chemistry Class 12 - NEET PDF Download

Who discovered Raoult's Law?

• Raoult’s law has been named after François-Marie Raoult, a French chemist who while experimenting found out that when substances were mixed in a solution, the vapour pressure of the solution decreased simultaneously. 
• Raoult’s law was established in 1887  and is also considered the law of thermodynamics.

What is Raoult’s Law?

Raoult’s law states that:

For a solution of volatile liquids, the partial vapour pressure of each component of the solution

is directly proportional to its mole fraction present in solution.

Mathematical representation of Raoult’s law:

Let's understand this by looking at the example below

Ideal Solution of Two Volatile Liquids

Let two volatile liquids A and B dissolve each other to form an ideal solution. Then the vapours above the solution will contain the vapours of A and B. Applying Dalton’s law of partial pressure the total vapour pressure of the solution will be 

P_solution = P_A + P_B

Where,
 P_A and P_B are the partial vapour pressures of A and B.

P_A = P^º_A X_A

P_B = P^º_B X_B

Where:

• P^º_A and P^º_B are vapour pressures of pure A and B  respectively
• x_A and x_B are mole fractions of A and B in a liquid solution respectively

Hence, 𝑃 (Solution) 𝑜𝑟 P_s =   P_A⁰ X_A + P_B⁰ X_B

Also, P_s = P⁰_A(1-X_B) + P⁰_BX_B

⇒ P_s = P⁰_A+(P⁰_B - P⁰_A)X_B

This equation is of the form y = mx + c. The value of m (slope) may be (+ve) or (–ve) depending upon whether P⁰_B > P⁰_A (m = + ve) or P⁰_B < P⁰_A (m = –ve).

Hence a plot of a graph of P_s versus X_B will be a straight line with an intercept on the y-axis equal to p_A and a slope equal to (P⁰_B - P⁰_A)

Plot of Vapour Pressure and Mole Fraction 

The total Pressure of a Solution will be:

• P (Solution) = PA + PB = x_AP⁰_A + x_BP⁰_B
  

Finding Vapour Composition

• Vapour composition means to find out the mole fraction of A and B in vapour i.e y_A and y_B. 
• Dalton’s law of particle pressure is used to calculate the vapour composition, as we know, Partial pressure = mole fraction × total pressure
  ⇒ p_A= y_A × p_S
  or y_A =𝑃_A/𝑃_𝑆
  Similarly for B
  ⇒ y_B = 𝑃_A/𝑃_𝑆

where y_A and y_B are mole fractions of ‘A’ and ‘B’ in the vapour phase and y_A +  y_B = 1

SPECIAL NOTE: 
x_A and x_B are mole fraction of ‘A’ and ‘B’ in the solution phase and x_A +  x_B =1

y_A and y_B are mole fraction of ‘A’ and ‘B’ in the vapour phase and y_A +  y_B =1

Raoult's Law as a special case of Henry's law

• At a given temperature liquids vaporize. At equilibrium, the pressure exerted by the vapour of the liquid over the liquid phase is referred to as vapour pressure.
• According to Raoult’s law, the vapour pressure of a volatile component in a given solution can be defined by p_i = p_i˚x_i
• In an answer of a gas in a fluid one of the segments is volatile to the point that it exists as a gas and solvency is given by Henry's law which expresses that p = K_H x
• Comparing both the equations we get that partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. The proportionality constant K_H differs from p_i˚
• Therefore, Raoult's law turns into a unique instance of Henry’s law in which K_H get to be equivalent to p_i˚

Limitations of Raoult’s Law

There are a few limitations to Raoult’s law:

1. Raoult’s law is apt for describing ideal solutions. However, ideal solutions are hard to find and they are rare. Different chemical components have to be chemically identical equally.
2. Since many of the liquids that are in the mixture do not have the same uniformity in terms of attractive forces, these type of solutions tends to deviate away from the law.

There is either a negative or a positive deviation: 

• Negative deviation happens when the vapour pressure is lower than what Raoult's law predicts. For instance, this occurs in a mix of chloroform and acetone or in a solution of water and hydrochloric acid.
• On the other hand, positive deviation occurs when the attraction among similar molecules is stronger, surpassing the attraction between dissimilar molecules. In such cases, both components of the mixture can easily evaporate from the solution. An example of positive deviation is seen in mixtures like benzene and methanol or ethanol and chloroform.

Positive and Negative Deviation

Solved Examples

Example 1. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate the total vapour pressure of the solution.
Solution. 
Number of moles of ethanol = 60/40 = 1.5
Number of moles of methanol = 40/32 = 1.25
X_A = 1.25/1.25 + 1.3 = 0.4545 and X_B = 1 – 0.4545 = 0.545
Let A = CH₃OH, B = C₂H₅OH
Total pressure of the solution
P_T = X_AP⁰_A + X_BP⁰B
= 0.4545 × 88.7 + 0.545 × 44.5 = 40.31 + 24.27
= 64.58 mm Hg

Example 2. The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If X_A and Y_A are the mole fraction of A in the liquid and vapour, respectively find the value of X_A for which Y_A − X_A has a minimum. What is the value of the pressure at this composition?
Solution. 

Subtracting x_A from both sides, we get

Now differentiating w.r.t. x_A, we get

The value of x_A at which y_A − x_A has a minimum value can be obtained by putting the above derivative equal to zero. Thus we have

Solving for x_A, we get 

 
hence


Example 3. A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm³ of benzene (density 0.889 g cm^-3). At room temperature, the vapour pressure of this solution is 99.88 mm Hg while that of benzene is 100mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degrees lower than that of benzene. What is the value of the molal freezing point depression constant of benzene?
Solution.





Also,
ΔT_f = K_f × molality f
∴ 0.73 = K × 0.1452
f_K = 5.028 K. molality

Example 4. What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution with a mole fraction of benzene of 0.400? 
(P^o_B = 119 torr and P_T^o = 37.0 torr)
Solution.
The Total pressure of the solution is given by
P_T = X_B P⁰_B + X_TP⁰_T
= 0.4 × 119 + 0.6 × 37
= 47.6 + 22.2
= 69.8 torr
Applying Dalton’s law for mole fraction in the vapour phase.
Y_B = P_B/P_T = 47.6/62.4
= 0.763
Y_T = 1– 0.763
= 0.237