Q.[Note: two or more options may be correct] Intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

The key idea is the characteristics of ideal solutions.

The problem states that intermolecular forces between benzene-benzene and toluene-toluene are nearly of the same strength. This indicates that the intermolecular forces between benzene-toluene molecules will also be comparable, leading to the formation of an ideal solution.

1. For an ideal solution, there is no net change in enthalpy or volume upon mixing. Therefore, $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. This means options (i) and (ii) are true statements. …

When two liquids mix, the nature of the intermolecular forces between the molecules plays a crucial role in determining the properties of the resulting solution. An ideal solution is a theoretical construct where the interactions between different types of molecules (A-B) are identical in strength to the interactions between molecules of the same type (A-A and B-B).

This condition of identical intermolecular forces has direct consequences for the thermodynamic properties of mixing:

• Enthalpy of mixing ($\Delta_{mix}H$): If the forces are identical, there is no net energy change when molecules rearrange from pure components to a mixture. No heat is absorbed or released. Thus, $\Delta_{mix}H = 0$.
• Volume of mixing ($\Delta_{mix}V$): Similarly, if the forces are identical, there is no change in the average distance between molecules upon mixing. The total volume of the solution is simply the sum of the volumes of the pure components. Thus, $\Delta_{mix}V = 0$.
• Raoult's Law: Ideal solutions strictly obey Raoult's Law over the entire range of concentrations.
• Azeotropes: Ideal solutions do not form azeotropes. Azeotropes are constant-boiling mixtures that form when there are significant positive or negative deviations from Raoult's Law.

The problem states that intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. Benzene and toluene are also structurally very similar (toluene is methylbenzene). This similarity in structure and intermolecular forces strongly suggests that a mixture of benzene and toluene will behave as a nearly ideal solution.

Let's evaluate each option based on this understanding:

1. Analyze the given information:

   The core information is that intermolecular forces between benzene-benzene (let's call them A-A) are nearly the same strength as those between toluene-toluene (B-B). Given their structural similarity, it is also reasonable to assume that the intermolecular forces between benzene and toluene (A-B) molecules will be very similar to A-A and B-B interactions. This is the defining characteristic of an ideal solution.

2. Evaluate option (i) $\Delta_{mix}H = zero$:

   For an ideal solution, there is no net change in the energy of intermolecular interactions upon mixing. The energy required to break A-A and B-B interactions is compensated by the energy released in forming A-B interactions. Therefore, the enthalpy of mixing is zero.

   For an ideal solution, $\Delta_{mix}H = 0$.

   Since benzene and toluene form a nearly ideal solution, this statement is true.

3. Evaluate option (ii) $\Delta_{mix}V = zero$:

   In an ideal solution, there is no change in the volume occupied by the molecules when they mix. The molecules fit together perfectly without any expansion or contraction. Therefore, the volume of mixing is zero. …

Concept: Ideal Solutions and Azeotropes

Method: Raoult's Law and Thermodynamic Mixing Properties

Step 1: Identify the nature of the mixture

Benzene and toluene have nearly identical intermolecular forces (both are non-polar aromatic hydrocarbons). This means:

• $A-A$, $B-B$, and $A-B$ interactions are approximately equal.
• The solution is expected to obey Raoult's Law over the entire composition range.

Step 2: Recall properties of an ideal solution

For an ideal solution:

• $\Delta_{mix}H = 0$ (no heat absorbed or released)
• $\Delta_{mix}V = 0$ (no volume change)
• No azeotrope formation
• Vapor pressure follows Raoult's Law

Step 3: Check each option

Concept in brief

• Benzene and toluene are structurally similar (both aromatic hydrocarbons).
• Their intermolecular forces (dispersion forces) are nearly equal.
• Therefore, when mixed:
  • No net energy change → $\Delta_{mix}H \approx 0$
  • No net volume change → $\Delta_{mix}V \approx 0$
  • They obey Raoult’s law → form an ideal solution.
• An ideal solution does not form an azeotrope (neither minimum nor maximum boiling).

Common mistakes & how to avoid them

✗ Mistake 1: Thinking “nearly same strength” means “not ideal”

• Why it happens: Students see “nearly” and assume it’s not exact, so they think the solution is non-ideal.
• How to avoid: In exam language, “nearly of same strength” means the difference is negligible. For ideal solution criteria, we only need $\Delta_{mix}H \approx 0$ and $\Delta_{mix}V \approx 0$. That’s satisfied here. → They form an ideal solution. So option (iv) is not true (it says they will not form ideal solution — that’s false).

✗ Mistake 2: Confusing azeotropes with ideal solutions

• Why it happens: Many students think “minimum boiling azeotrope” is a property of all solutions with similar forces.
• How to avoid: Remember:
  • Ideal solutions → no azeotrope.
  • Azeotropes arise only from non-ideal behavior (large deviations from Raoult’s law). Since benzene + toluene is ideal, option (iii) (“These will form minimum boiling azeotrope”) is not true.

✗ Mistake 3: Assuming $\Delta_{mix}H$ and $\Delta_{mix}V$ are always zero for any mixture

• Why it happens: Students memorize “ideal solution → $\Delta_{mix}H = 0$, $\Delta_{mix}V = 0$” but forget the reason.
• How to avoid: …