Question

Based on the information given below, mark the correct option. Information:

(i) In bromoethane and chloroethane mixture intermolecular interactions of $A–A$ and $B–B$ type are nearly same as $A–B$ type interactions

(ii) In ethanol and acetone mixture $A–A$ or $B–B$ type intermolecular interactions are stronger than $A–B$ type interactions.

(iii) In chloroform and acetone mixture $A–A$ or $B–B$ type intermolecular interactions are weaker than $A–B$ type interactions.

• A. Solution (ii) and (iii) will follow Raoult’s law.
• B. Solution (i) will follow Raoult’s law.
  
  
• C. Solution (iii) will show positive deviation from Raoult’s law
• D. Solution (ii) will show negative deviation from Raoult’s law.
  

Answer 1

The correct option is B Solution (i) will follow Raoult’s law.



Raoult’s law states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Mathematically, the equation for Raoult’s law is:

$P_A=X_A{P}_A^0$

Where,

$P_A$ = partial pressure of component A

$X_A$ = mole fraction of component A

$P_A^O$ = vapour pressure of the pure component A

Ideal solutions: The solutions that obey Raoult’s law are called ideal solutions.

In these solutions, the intermolecular attractive forces present between the solute and the solvent molecules are nearly equal to those present between the solute molecules and the solvent molecules individually.

Non-ideal solutions: The solutions that do not obey Raoult’s law are called non-ideal solutions.

In these solutions, the intermolecular attractive forces present between the solute and the solvent molecules are not equal to those present between the solute molecules and the solvent molecules individually.

Solution (i)

In bromoethane-chloroethane mixture, the intermolecular interactions of the $A–A$ and $B–B$ types are nearly the same as the $A–B$ type interaction, which means this solution follows Raoult’s law.

Solution (ii)

In its pure state, ethanol molecules are strongly bonded through hydrogen bonding, and when acetone is added to this solution, acetone molecules get in between the molecules of ethanol and breaks some of the hydrogen bonds.

Thus, in ethanol-acetone mixture, the $A–A$ or $B–B$ type intermolecular interactions are stronger than the $A–B$ type interactions. So, this solution is a non-ideal solution showing positive deviation from Raoult’s law.

Solution (iii)

When chloroform and acetone solutions are mixed, hydrogen bonds are formed between them, which increases the intermolecular interactions.

Thus, in chloroform-acetone mixture, the $A–A$ or $B–B$ type intermolecular interactions are weaker than the $A–B$ type interaction. So, this solution is a non-ideal solution showing negative deviation from Raoult’s law.

Thus, only solution (i) will follow Raoult’s law.

Hence, option (B) is correct.