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

All you need of Class 12 at this link: Class 12

**RAOULT’S LAW**

- 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. - 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} - 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 y-axis equal to p_{A}and slope equal to (P^{0}_{B}- P^{0}_{A})

**Total Pressure of a Solution will be:**

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

**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! =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 vapour phase and y_{A}+ y_{B}= 1

**SPECIAL NOTE**

- x
_{A}and x_{B}are mole fraction of ‘A’ and ‘B’ in solution phase and

o x_{A}+ x_{B}=1 - y
_{A}and y_{B}are mole fraction of ‘A’ and ‘B’ in vapour phase and

o 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, vapour pressure of a volatile component in a given solution ca 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}˚

**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 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 degree lower than that of benzene. What is the value of 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 vapour phase.

Y

= 0.763

Y

= 0.237

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

255 videos|306 docs|201 tests

### Vapour Solutions and Types of Solutions

- Doc | 7 pages
### Test: Solubility

- Test | 10 ques | 15 min
### Fun Video: What are the Solutions ?

- Video | 08:21 min
### Test: Solutions 1 - From Past 28 Years Questions

- Test | 17 ques | 35 min
### NCERT Textbook - Solutions

- Doc | 30 pages
### Short & Long Answer Question (Part - 2) - Solutions

- Doc | 8 pages

- Solved Examples - Solutions (Part - 1)
- Doc | 4 pages
- Previous year Questions (2016-20) - Solutions
- Doc | 12 pages