Q.The principle involved in paper chromatography is

Concept & Intuition

Paper chromatography is often misunderstood as “adsorption” because the paper looks solid. But the key insight is that the paper fibers hold water (adsorbed from humidity or added deliberately), and it is this water layer that acts as the stationary phase. The sample components then partition (dissolve) between the water and the moving solvent — exactly like liquid-liquid extraction. So it’s partition, not adsorption. The paper itself is just an inert support.

1. Statement I: “It is a type of adsorption chromatography”

   Adsorption chromatography relies on the sample sticking to a solid surface (e.g., silica gel in TLC). In paper chromatography, the sample does not primarily stick to the cellulose fibers; it distributes between two liquids. Hence, I is false.

2. Statement II: “It is a type of partition chromatography”

   Correct. The stationary phase is a thin film of water (or other liquid) held on the paper. The mobile phase (solvent) moves by capillary action, and the sample partitions between the two liquid phases. II is true.

3. Statement III: “Both stationary and mobile phases are liquids”

   The mobile phase is a liquid solvent. The stationary phase is also a liquid (water) — even though it’s supported on a solid. So III is true.

4. Statement IV: “Stationary phase is solid and mobile phase is liquid” …

The question tests your understanding of how the physical properties of components in a mixture determine the best separation method. You don’t memorise pairings — you reason from the property.

1. Mixture A: $\mathrm{CHCl_3}$ (b.p. 334 K) + $\mathrm{C_6H_5NH_2}$ (b.p. 457 K)

   The boiling points differ by over 120 K. When the difference is large (typically > 30–40 K) and neither component decomposes near its boiling point, simple distillation works cleanly: the lower-boiling chloroform distils over first, leaving aniline behind.

   → A matches II (Distillation).

2. Mixture B: Naphthalene + NaCl

   Naphthalene is a volatile solid that sublimes (turns directly from solid to vapour) at moderate heating, while NaCl is a non-volatile ionic solid that does not. Heating the mixture lets naphthalene vapour rise and re‑deposit as pure crystals on a cold surface — this is sublimation.

   → B matches III (Sublimation).

3. Mixture C: $\mathrm{C_6H_5NH_2}$ + $\mathrm{H_2O}$

   Aniline is only slightly soluble in water and forms a heterogeneous mixture. More importantly, aniline and water form a minimum-boiling azeotrope that boils below the boiling point of either pure component. Steam distillation is the classic method to separate aniline from water: steam carries the volatile aniline vapour, which then condenses as two immiscible layers that can be separated.

   → C matches I (Steam distillation).

4. Mixture D: Organic compound present in aqueous medium …

Concept & Intuition

This is a classic metallurgy recall problem. Each statement checks a specific fact from extraction processes. The trick is to know the exact composition of copper matte — a common pitfall is confusing CuS (copper(II) sulphide) with Cu₂S (copper(I) sulphide). Froth flotation is indeed the standard method for concentrating sulphide ores because sulphide particles are preferentially wetted by oil-based frothers. Distillation works for metals like zinc or mercury that boil at relatively low temperatures.

Step-by-step reasoning

1. Statement I: “Distillation method is used for refining low boiling metals”

   • Distillation relies on differences in boiling points. Metals with low boiling points (e.g., zinc, cadmium, mercury) can be vaporized and then condensed separately from impurities.
   • This is a standard refining technique for such metals.
   • Conclusion: Statement I is correct.

2. Statement II: “Copper matte contains CuS and FeS”

   • In the extraction of copper, the smelting of roasted ore produces a molten mixture called matte.
   • The actual composition is Cu₂S (copper(I) sulphide) and FeS (iron(II) sulphide). CuS (copper(II) sulphide) is not stable at smelting temperatures; it decomposes to Cu₂S.
   • The statement says “CuS” — that is chemically incorrect.
   • Conclusion: Statement II is false.

3. Statement III: “Froth floatation method is used in concentration of sulphide ores” …

The question tests your understanding of how the choice of refining process depends on the metal’s melting point, volatility, reactivity, and the nature of impurities. Each method exploits a specific difference between the metal and its impurities.

Let’s go through each metal one by one.

1. Nickel (Ni) — This is a high-melting transition metal. It forms a volatile carbonyl compound, $\text{Ni(CO)}_4$, when reacted with carbon monoxide. This compound decomposes on heating to give pure nickel. This is vapour phase refining (the Mond process). So A matches with V.

2. Silicon (Si) — Silicon is used in semiconductors and requires extreme purity. Its impurities (like boron or phosphorus) have slightly different solubilities in solid vs. liquid silicon. Zone refining uses a moving heater to sweep impurities to one end of a rod, leaving the rest ultrapure. So B matches with I.

3. Tin (Sn) — Tin has a low melting point. Impurities like iron or copper have much higher melting points. In liquation, the impure metal is heated gently; tin melts and flows away, leaving the solid impurities behind. So C matches with II. …

The key idea here is what "deionized" actually means. Hard water contains dissolved salts — these are ionic compounds that split into positive ions (cations like $\text{Ca}^{2+}$, $\text{Mg}^{2+}$, $\text{Na}^+$) and negative ions (anions like $\text{Cl}^-$, $\text{SO}_4^{2-}$, $\text{HCO}_3^-$). To get truly deionized water, you must strip away all of these ions, not just the ones that cause hardness.

Zeolites (option A) are used in water softening — they swap hardness-causing cations for sodium ions. But the water still has sodium ions and all the original anions, so it's softened, not deionized. That's a different goal.

Now let's walk through the logic step by step.

1. A single ion exchanger only handles one charge type. A cationic exchanger has acidic groups (like $-\text{SO}_3\text{H}$) that swap $\text{H}^+$ for metal cations. An anionic exchanger has basic groups (like $-\text{NH}_2$) that swap $\text{OH}^-$ for anions. Neither alone can remove both kinds of ions.

2. If you use only a cationic exchanger, the water leaving it contains $\text{H}^+$ (from the exchanger) plus all the original anions. That's acidic water, not deionized water. Same problem in reverse with only an anionic exchanger — you'd get basic water with leftover cations.

3. The correct sequence is: pass water first through a cationic exchanger, then through an anionic exchanger. In the first column, all cations are replaced by $\text{H}^+$ ions. In the second column, all anions are replaced by $\text{OH}^-$ ions. The $\text{H}^+$ and $\text{OH}^-$ then combine to form pure water: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$. …

The question asks us to match several key terms from the study of solutions and gases to their correct definitions or associated phenomena. This requires a clear understanding of colligative properties, which depend on the number of solute particles, and fundamental gas laws.

1. Understanding Ebullioscopic Constant:

   The ebullioscopic constant, denoted as $K_b$, is a proportionality constant that relates the molality of a solute to the elevation of the boiling point of a solvent. When a non-volatile solute is added to a solvent, the boiling point of the solution increases compared to the pure solvent. This phenomenon is called elevation of boiling point.

   The elevation of boiling point ($\Delta T_b$) is given by $\Delta T_b = K_b \cdot m$, where $m$ is the molality of the solution.

   Therefore, the ebullioscopic constant is directly associated with the elevation of boiling point.

   • Match: A) Ebullioscopic constant $\rightarrow$ III) Elevation of boiling point

2. Understanding Cryoscopic Constant:

   The cryoscopic constant, denoted as $K_f$, is a proportionality constant that relates the molality of a solute to the depression of the freezing point of a solvent. When a non-volatile solute is added to a solvent, the freezing point of the solution decreases compared to the pure solvent. This phenomenon is called depression of freezing point.

   The depression of freezing point ($\Delta T_f$) is given by $\Delta T_f = K_f \cdot m$, where $m$ is the molality of the solution.

   Therefore, the cryoscopic constant is directly associated with the depression of freezing point.

   • Match: B) Cryoscopic constant $\rightarrow$ I) Depression of freezing point

3. Understanding Henry's Law:

   Henry's law describes the relationship between the partial pressure of a gas above a liquid and the solubility of that gas in the liquid. It states that the partial pressure of the gas in the vapor phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution.

   Henry's law is expressed as $p = K_H x$, where $K_H$ is Henry's law constant.

   This law is fundamental to understanding how gases dissolve in liquids, such as oxygen dissolving in water or carbon dioxide in soft drinks.

   • Match: C) Henry’s law $\rightarrow$ IV) Solubility of a gas in liquid …