Question

The intermolecular forces between two benzene molecules are of nearly the same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

• A. These will form minimum boiling azeotrope.
  
  
• B. ${\Delta }_{mix}V=\text{zero}$
  
  
• C.
  These will not form an ideal solution
• D. ${\Delta }_{mix}H=\text{zero}$
  
  

Answer 1

Answer: (A), (C)

These will not form an ideal solution

Ideal solution

A solution of benzene and toluene constitutes an ideal solution. An ideal solution follows Raoult’s law over the entire range of concentrations.

Raoult’s law states that “for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution.”

Ideal solutions have the following two major properties :

1. Enthalpy of mixing of the pure components to form the solution is zero. $({\Delta }_{mix}H=\text{zero})$
2. The volume of mixing is also zero.$({\Delta }_{mix}V=\text{zero})$.

Analysing the options :

$\left(A\right)$ ${\Delta }_{mix}H=\text{zero}$.So, the option A is true.

$\left(B\right)$ ${\Delta }_{mix}V=\text{zero}$.So, the option B is true.

$\left(C\right)$ The solution formed is ideal. Therefore, it does not show positive or negative deviation from Raoult's law. Hence, it does not form an azeotrope. Thus, option (c) is not true about the solution .

$\left(D\right)$ The solution formed is ideal. So, option (D) is not true about the solution.

Hence, options (C) and (D) are the correct answers.