Q.Give an example of a solid solution in which the solute is a gas.

Concept and Intuition

An azeotrope is a mixture that boils at a constant composition because the vapour has the same composition as the liquid. For a maximum‑boiling azeotrope, the boiling point of the mixture is higher than that of either pure component. This happens when the intermolecular attractions between the two different molecules are stronger than those in the pure liquids — a situation called negative deviation from Raoult’s law. The stronger cross‑attractions make it harder for molecules to escape into the vapour, lowering the vapour pressure and raising the boiling point.

The key is to identify which pair forms such strong cross‑interactions. Let’s examine each option.

Step‑by‑Step Reasoning

1. Option (A): Benzene‑toluene

   Benzene and toluene are both non‑polar hydrocarbons with similar structures. Their intermolecular forces (dispersion forces) are nearly identical, so the mixture behaves almost ideally — no significant deviation from Raoult’s law. No azeotrope forms; the boiling point lies between those of the pure components.

   → Not a maximum‑boiling azeotrope.

2. Option (B): Carbon disulphide‑acetone

   Carbon disulphide ($\mathrm{CS_2}$) is non‑polar, while acetone ($\mathrm{CH_3COCH_3}$) is polar. Mixing them disrupts the dipole‑dipole interactions in acetone, making it easier for molecules to escape. This gives a positive deviation from Raoult’s law (higher vapour pressure, lower boiling point). Such systems form minimum‑boiling azeotropes, not maximum‑boiling ones.

   → Incorrect.

3. Option (C): Ethyl alcohol‑water

   Ethanol and water both form strong hydrogen bonds. However, when mixed, the hydrogen‑bond network in each pure liquid is partially broken, and the cross‑hydrogen bonds (ethanol‑water) are weaker than the sum of the pure‑liquid interactions. This leads to a positive deviation (higher vapour pressure, lower boiling point). Ethanol‑water indeed forms a minimum‑boiling azeotrope (boiling point ~78.2 °C, lower than both pure ethanol and water).

   → Incorrect.

4. Option (D): Acetone‑chloroform

   Acetone ($\mathrm{CH_3COCH_3}$) has a carbonyl oxygen with lone pairs; chloroform ($\mathrm{CHCl_3}$) has a hydrogen attached to three electronegative chlorines. The hydrogen in chloroform is sufficiently acidic to form a hydrogen bond with the oxygen of acetone:

This cross‑hydrogen bond is stronger than the dipole‑dipole interactions in pure acetone or pure chloroform. As a result, the vapour pressure of the mixture is lower than predicted by Raoult’s law (negative deviation), and the boiling point is higher than either pure component. The system forms a maximum‑boiling azeotrope at about 64.5 °C (pure acetone boils at 56 °C, pure chloroform at 61 °C).

→ Correct.

Concept & Intuition

Benzene and toluene form an ideal solution (Raoult’s law holds). At a given temperature, benzene has a higher vapour pressure than toluene — it is more volatile. When a liquid mixture evaporates, the vapour is richer in the more volatile component. This is the key idea behind distillation: the vapour composition differs from the liquid composition. Here, with equal moles in the liquid, the vapour will not be 50:50; it will be skewed toward benzene.

Step-by-step reasoning

1. Identify the vapour pressures

   At 25 °C, pure benzene has vapour pressure $P_B^0 \approx 95.1$ mm Hg, and pure toluene has $P_T^0 \approx 28.4$ mm Hg. (Exact values depend on temperature, but the key is $P_B^0 > P_T^0$.)

2. Apply Raoult’s law for the liquid mixture

   For 1 mole each, mole fractions in the liquid are $x_B = 0.5$ and $x_T = 0.5$.

   Partial pressures above the liquid:

3. Find the vapour composition The total pressure $P_{\text{total}} = P_B + P_T = 61.75$ mm Hg. Mole fraction of benzene in the vapour ($y_B$) is given by Dalton’s law:

So the vapour is about 77 % benzene and 23 % toluene.

4. Interpret the result Even though the liquid had equal moles, the vapour contains a higher percentage of benzene. This is a general property: the vapour is always enriched in the more volatile component.

Concept & Intuition

Raoult’s law states that for an ideal solution, the partial vapour pressure of each component equals its mole fraction times its pure vapour pressure. The total vapour pressure is the sum of these partial pressures. Here we have two different compositions (before and after adding extra X), each giving a total pressure. That gives two linear equations in the two unknowns $P_X^0$ and $P_Y^0$. Solving them is straightforward — no tricks, just careful algebra.

Step-by-step solution

1. Initial condition Moles of X = 1, moles of Y = 1 → total moles = 2. Mole fractions:

Total vapour pressure $P_{\text{total}} = 1000$ mm.

By Raoult’s law:

Multiply by 2:

2. After adding 2.0 moles of X New moles: X = 1 + 2 = 3, Y = 1 (unchanged) → total moles = 4. New mole fractions:

Vapour pressure increases by 200 mm, so new total pressure = $1000 + 200 = 1200$ mm.

Raoult’s law gives:

Multiply by 4:

3. Solve the system Subtract Equation 1 from Equation 2:

Substitute into Equation 1:

Concept & Intuition

For an ideal solution of two volatile liquids, the vapour above the solution is not simply the same composition as the liquid. Instead, the more volatile component (higher pure vapour pressure) enriches the vapour. Raoult’s law gives the partial pressure of each component in the vapour:

where $x$ is the mole fraction in the liquid and $P^\circ$ is the pure vapour pressure. Dalton’s law then tells us the total pressure is the sum of partial pressures, and the mole fraction of a component in the vapour is its partial pressure divided by the total pressure. So we can compute the vapour composition directly from the liquid composition and pure vapour pressures.

Step-by-step solution

1. Identify given data

   • Pure vapour pressures: $P_X^\circ = 100\ \text{mm Hg}$, $P_Y^\circ = 200\ \text{mm Hg}$
   • Mole fraction of X in the liquid: $x_X = 0.4$
   • Since it’s a binary solution, $x_Y = 1 - x_X = 0.6$
   • Mole fraction of Y in the vapour is given as $\frac{a}{20}$.

2. Apply Raoult’s law to find partial pressures

3. Find total vapour pressure

4. Compute mole fraction of Y in the vapour By Dalton’s law, the mole fraction of Y in the vapour phase is:

5. Relate to the given expression The problem states $y_Y = \frac{a}{20}$. Therefore:

Concept & Intuition

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lower than that of the pure solvent. Raoult’s law tells us that the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute. Here, the vapour pressure is reduced to 60% of its original value — meaning the lowering is 40% of the original. That fraction directly gives the mole fraction of solute, and from there we can find its molar mass.

Step-by-step solution

1. Interpret the vapour pressure reduction Let $P^0$ be the vapour pressure of pure xylene. The problem says the vapour pressure is reduced to $60\%$ of $P^0$. So the new vapour pressure $P = 0.60\,P^0$. The lowering of vapour pressure is:

Hence the relative lowering is:

2. Apply Raoult’s law for a non-volatile solute For a dilute solution of a non-volatile solute, Raoult’s law states:

where $x_{\text{solute}}$ is the mole fraction of the solute. Therefore:

3. Express mole fraction in terms of masses and molar masses

   Let $M_A$ be the molar mass of solute A (in g/mol).

   Mass of solute = 60 g → moles of solute = $\frac{60}{M_A}$.

   Mass of solvent (xylene) = 212 g, molar mass of xylene = 106 g/mol → moles of solvent = $\frac{212}{106} = 2.00$ mol.

   The mole fraction of solute is:

4. Set up the equation and solve We know $x_{\text{solute}} = 0.40$, so:

Multiply both sides by the denominator:

Subtract $\frac{24}{M_A}$ from both sides:

Hence:

Minimum-boiling azeotropes form in solutions showing positive deviation (weaker A–B interactions than A–A/B–B, higher vapour pressure, lower boiling point).

• (A) Phenol–aniline: strong O–H···N hydrogen bonding between components → negative deviation → maximum-boiling.
• (B) Ethyl alcohol–water: positive deviation → minimum-boiling azeotrope (≈95.6% ethanol by mass).
• (C) Acetone–chloroform: new H-bond (C–H···O) between them → negative deviation → maximum-boiling.
• (D) Nitric acid–water: negative deviation → maximum-boiling azeotrope (≈68% HNO$_3$).

Negative deviation from Raoult's law $\Rightarrow$ observed vapour pressure lower than ideal, arising when solute–solvent attractions exceed solvent–solvent and solute–solute attractions.

Evaluating the mixtures:

• Benzene–acetone: breaks acetone's dipole interactions; weaker new forces $\Rightarrow$ positive deviation.
• Benzene–chloroform: chloroform's acidic C–H forms a weak attractive interaction with the benzene $\pi$-cloud, giving stronger A–B forces $\Rightarrow$ negative deviation.
• Benzene–ethanol: disrupts ethanol H-bonding $\Rightarrow$ positive deviation.
• Benzene–$\mathrm{CCl_4}$: similar non-polar molecules $\Rightarrow$ nearly ideal / slight positive deviation.

Hence benzene–chloroform is the negative-deviation system.

Concept: Azeotropes. A minimum-boiling azeotrope is formed by solutions showing a LARGE POSITIVE deviation from Raoult's law; a maximum-boiling azeotrope by solutions showing a large negative deviation.

1. Heptane + octane: chemically very similar, near-ideal solution - strictly no azeotrope at all.
2. Water + nitric acid: strong H-bonding/interaction -> negative deviation -> MAXIMUM boiling azeotrope (not minimum).
3. Ethanol + water: the textbook example of a positive deviation -> MINIMUM boiling azeotrope (95% ethanol, 351.1 K). YES.
4. Acetone + carbon disulphide (printed here as 'carbon dioxide' - the standard NCERT pair is acetone + CS2): positive deviation -> MINIMUM boiling azeotrope. YES. So the genuine minimum-boiling pairs are (iii) and (iv). Option (ii) is definitely excluded (it is maximum boiling), which rules out options (A) and (B). Option (D) omits ethanol + water, the canonical minimum-boiling azeotrope. The only option containing both (iii) and (iv) is (C).
   ✓Final answer

   The correct option is (C) — (i), (iii), (iv)

   ANSWER: C

Concept: Azeotropes form only from NON-IDEAL solutions (those showing appreciable positive or negative deviation from Raoult's law). Ideal solutions never form azeotropes.

• Chlorobenzene + bromobenzene: nearly ideal (very similar structures) - no azeotrope
• C2H5Br + C2H5Cl: nearly ideal - no azeotrope
• n-hexane + n-heptane: classic ideal solution - no azeotrope
• CCl4 + CHCl3: shows a measurable (positive) deviation from Raoult's law because CHCl3-CHCl3 / CCl4-CCl4 interactions differ from the mixed ones; this pair is the non-ideal one and forms an azeotropic mixture.

ANSWER: D

Step 1 — The rule behind Raoult's-law deviations.

For a solution of A and B, let $A\!-\!B$ be the new solute–solvent interaction and $A\!-\!A$, $B\!-\!B$ the interactions in the pure liquids.

Step 2 — Solution A: acetone + chloroform.

Chloroform's C–H bond is strongly polarised by its three electronegative chlorine atoms, so its hydrogen is acidic enough to form a genuine hydrogen bond with the lone pair on the carbonyl oxygen of acetone:

This $A\!-\!B$ attraction is stronger than the dipole–dipole forces present in either pure liquid. The molecules are therefore held back from the vapour phase, the total vapour pressure falls below the Raoult prediction ⇒ negative deviation (this pair also forms a maximum-boiling azeotrope).

Step 3 — Solution B: acetone + carbon disulphide.

$CS_2$ is a linear, non-polar molecule (its two C=S bond dipoles cancel), so it can offer only weak dispersion forces to acetone. Mixing it with acetone breaks the relatively strong dipole–dipole attractions between acetone molecules and replaces them with much weaker acetone–$CS_2$ interactions. Molecules escape more easily, the vapour pressure exceeds the Raoult prediction ⇒ positive deviation (this pair forms a minimum-boiling azeotrope).

Step 4 — Read off the order asked.

Solution A ⇒ negative; Solution B ⇒ positive.

ANSWER: D