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# Raoult's Law Class 12 Notes | EduRev

## Class 12 : Raoult's Law Class 12 Notes | EduRev

The document Raoult's Law Class 12 Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
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Who discovers Raoult's Law?

• 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:

Psolution = Χsolvent P0solvent

Where,

Psolution = vapour pressure of the solution

Χsolvent  = mole fraction of the solvent

P0solvent = 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

Psolution = PA + PB

Where,
PA and PB 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

PA = PºA XA
PB = PºB XB

Where:

• PºA and PºB are vapour pressures of pure A and B  respectively
• xA and xB are mole fraction of A and B in liquid solution respectively

Hence, 𝑃 (Solution) 𝑜𝑟 Ps =   PA0 XA + PB0 XB

Also, Ps = P0A(1-XB) + P0BXB

⇒ Ps = P0A+(P0B - P0A)XB

This equation is of the form y = mx + c. The value of m (slope) may be (+ve) or (–ve) depending upon whether P0B > P0A (m = + ve) or P0B < P0A (m = –ve).

Hence a plot of a graph of Ps versus XB will be a straight line with an intercept on the y-axis equal to pA and slope equal to (P0B - P0A)

Total Pressure of a Solution will be:
• P (Solution) = PA + PB = xAP0+ xBP0B Vapour Composition
• Vapour composition means to find out the mole fraction of A and B in vapour i.e yA and yB.
• Dalton’s law of particle pressure is used to calculate the vapour composition, as we know, Partial pressure = mole fraction × total pressure
⇒ pA= yA × pS
or yA =𝑃A/𝑃𝑆
Similarly for B
⇒ yB = 𝑃A/𝑃𝑆

where yA and yB are mole fraction of ‘A’ and ‘B’ in the vapour phase and yA +  yB = 1

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

yA and yB are mole fraction of ‘A’ and ‘B’ in the vapour phase and yA +  yB =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 pi = pi˚xi
• 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 = KH 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 KH differs from pi˚
• Therefore, Raoult's law turns into a unique instance of Henry’s law in which KH get to be equivalent to pi˚

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. 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= 1.25/1.25 + 1.3 = 0.4545 and XB = 1 – 0.4545 = 0.545
Let A = CH3OH, B = C2H5OH
Total pressure of the solution
PT = XAP0A + XBP0B
= 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 XA and YA are the mole fraction of A in the liquid and vapour, respectively find the value of XA for which YA − Xhas a minimum. What is the value of the pressure at this composition?
Solution. Subtracting xA from both sides, we get Now differentiating w.r.t. xA, we get The value of xA at which yA − xA has a minimum value can be obtained by putting the above derivative equal to zero. Thus we have Solving for xA, 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,
ΔTf = Kf × molality f
∴ 0.73 = K × 0.1452
fK = 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?
(PoB = 119 torr and PTo = 37.0 torr)
Solution.
Total pressure of the solution is given by
PT = XB P0B + XTP0T
= 0.4 × 119 + 0.6 × 37
= 47.6 + 22.2
= 69.8 torr
Applying Dalton’s law for mole fraction in the vapour phase.
YB = PB/PT = 47.6/62.4
= 0.763
YT = 1– 0.763
= 0.237
TRY YOURSELF!

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?

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

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

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## Chemistry Class 12

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