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**Who discovers **

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

**What is Raoult’s Law?**

Raoult’s law states that:

A solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

**Mathematically, Raoult’s law equation is written as:**

P_{solution} = Χ_{solvent }P^{0}_{solvent}

Where,

P_{solution} = vapour pressure of the solution

Χ_{solvent} = mole fraction of the solvent

P^{0}_{solvent} = vapour pressure of the pure solvent

**Let's understand this by looking at the example below:**

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**

Where,

P_{A} and P_{B} are the partial vapour pressures of A and B.

Solution being dilute (Condition for ideality) Raoult’s law can be applied.

**Raoult’s law for binary solutions state that **

The vapour pressure of any volatile constituent of a binary solution at any given temperature is equal to the product of the vapour pressure of pure constituent and its mole fraction

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 fraction of A and B in liquid solution respectively

Hence, 𝑃 (Solution) 𝑜𝑟 P_{s} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}

Also, P_{s} = P^{0}_{A}(1-X_{B}) + P^{0}_{B}X_{B}

⇒ P_{s} = P^{0}_{A}+(P^{0}_{B} - P^{0}_{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 P^{0}_{B} > P^{0}_{A} (m = + ve) or P^{0}_{B} < P^{0}_{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 slope equal to (P^{0}_{B} - P^{0}_{A})

- P (Solution) = PA + PB = x
_{A}P^{0}_{A }+ x_{B}P^{0}_{B}

- 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 fraction 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}=1y

_{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 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. 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:

- 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.
- 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. The negative deviation occurs when the vapour pressure is lower than expected from Raoult’s law. An example of negative deviation is a mixture of chloroform and acetone or a solution of water and hydrochloric acid.

Alternatively, positive deviation takes place when the cohesion between similar molecules is greater or that it exceeds adhesion between unlike or dissimilar molecules. Both components of the mixture can easily escape from the solution. An example of positive deviation includes the mixtures of benzene and methanol or ethanol and chloroform.

**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 = CH_{3}OH, B = C_{2}H_{5}OH

Total pressure of the solution

P_{T} = X_{A}P^{0}_{A} + X_{B}P^{0}B

= 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?**

Subtracting x

Now differentiating w.r.t. x

The value of x

Solving for x

hence**Example 3. A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm ^{3} 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?**

Also,

ΔT

∴ 0.73 = K × 0.1452

f

Total pressure of the solution is given by

P

= 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

= 0.763

Y

= 0.237

Try yourself:If a mixture of A and B boils at a temperature lower than the boiling point of either of the components, what kind of deviation does the mixture show?

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Try yourself:Considering a binary solution of components A and B obeys Raoult’s law, which of the following is true?

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Try yourself:“Total pressure of gas mixture is the sum of individual pressures”. Which law is reflected in this statement?

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