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Raoult’s Law - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • 23 Questions around this concept.

Solve by difficulty

Consider a binary mixture of benzene and toluene, which exhibits negative deviation from Raoult's Law.

Assertion: Non-ideal solutions showing negative deviation from Raoult's law have stronger intermolecular forces between unlike  molecules.

Reasoning: When two different substances are mixed, they may experience either positive or negative deviations from Raoult's law. In the case of negative deviation, the vapour pressure of the solution is lower than predicted by Raoult's law. This happens because the intermolecular forces between unlike molecules are stronger than those between like molecules. As a result, the attractive forces between the molecules in the solution are stronger, causing them to stick together more tightly and making it harder for them to escape into the vapour phase. Therefore, the assertion is true, and the reasoning provides a logical explanation for it.

A binary mixture of acetone and chloroform is found to exhibit negative deviation from Raoult's Law. The mole fraction of acetone in the mixture is 0.6, and the vapour pressure of pure acetone and pure chloroform are 350 mmHg and 200 mmHg, respectively. What is the vapour pressure of the mixture?

A binary liquid solution is prepared by mixing n - heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

Concepts Covered - 2

Non-Ideal Solution Showing Negative Deviation from Raoult's Law

Negative deviation is seen when total vapour pressure for any mole fraction is less than that expected from Raoult's law. This happens when the new interactions are stronger than the interactions in the pure components (A-B} > A-A or B-B interactions).

  • ΔH = -ve, ΔV = -ve
  • It forms maximum boiling azeotrope, for example, CHCl3 + CH3COCH3. For such solutions ΔV and ΔH are negative.

Example

  1. Chloroform + benzene
  2. Acetone + aniline
  3. Nitric acid HNO+ water
  4. Acetic acid + pyridine
  5. CH3OH + Acetone

For such solutions: ΔV and ΔH are negaative.
Here, (V.P)obs < (VP)exp
(B.P)obs > (B.P)exp
PA < PAoXA
PB < PBoXB
PT < PAoXA + PBoXB

Here ΔS has less value while ΔG has more -ve value and the values of other colligative properties increase.

Non-Ideal Solution Showing Positive Deviation from Raoult's Law

Positive deviation occurs when total vapour pressure for any mole fraction is more than what is expected according to Raoult's law. This happens when the new interactions are weaker than the interaction in the pure component (A-B < A-A or B-B interactions).

  • ΔH = +ve, ΔV = +ve
  • It forms minimum boiling azeotropes, for example, C2H5OH + cyclohexane. The H-bonding present in pure C2H5OH are cut off on adding cyclohexane.
    For such solution, ΔV and ΔH are positive.
    Here, (V.P)obs > (VP)exp
    (B.P)obs < (B.P)exp
    PA > PAoXA
    PB > PBoXB
    PT > PAoXA + PBoXB
  • Here Hydrogen bonds are broken or weakened. Here ΔS is more while ΔG has less -ve value and the values of other colligative properties also decrease.
  • Here minimum boiling point azeotrope mixture is formed.

Examples:

  1. C2H5OH + cyclohexane
  2. Acetone + carbon disulphide
  3. Acetone + benzene
  4. Acetone + Ethylalcohol
  5. Carbon tetrachloride + chloroform or Toluene
  6. Methylalcohol + water
  7. Water + Ethylalcohol

Study it with Videos

Non-Ideal Solution Showing Negative Deviation from Raoult's Law
Non-Ideal Solution Showing Positive Deviation from Raoult's Law

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