Step 1: Understand the types of defects.
A Schottky defect is a vacancy defect in crystalline solids where a pair of oppositely charged ions are missing from the lattice, maintaining overall electrical neutrality. It is common in highly ionic compounds where the cation and anion are of similar size (e.g., NaCl, KCl).
A Frenkel defect is a defect where an ion (usually the smaller cation) is displaced from its lattice position to an interstitial site. This is common in crystals with a large size difference between the cation and anion.

Step 2: Analyze the options.
- NaCl and KCl are classic examples of compounds showing Schottky defects.
- AgBr is a unique case that shows both Schottky and Frenkel defects. However, it is most commonly cited as an example of a Frenkel defect because the Ag\(^+\) ion is small and can easily move to an interstitial site. Therefore, among the given choices, AgBr is the compound where the Frenkel defect is more prominent and Schottky defect is not the primary defect observed. Quick Tip: To identify crystal defects: - {Schottky Defect:} Think "similar size" ions (e.g., Na\(^+\)/Cl\(^-\)). This defect decreases the density of the crystal. - {Frenkel Defect:} Think "large size difference" (e.g., small Ag\(^+\)/large Br\(^-\)). Density remains unchanged.

Step 1: Identify the formula for depression in freezing point.
The depression in freezing point (\(\Delta T_f\)) is calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \]
where \(i\) is the van't Hoff factor, \(K_f\) is the molal freezing point depression constant, and \(m\) is the molality of the solution.

Step 2: Determine the value of the van't Hoff factor (\(i\)).
Urea (\( CO(NH_2)_2 \)) is a non-electrolyte, meaning it does not dissociate into ions in solution. Therefore, its van't Hoff factor is \(i=1\).

Step 3: Calculate the depression in freezing point (\(\Delta T_f\)). \[ \Delta T_f = (1) \cdot (1.86 K kg mol^{-1}) \cdot (0.1 m) = 0.186 K or 0.186 \,^\circC \]

Step 4: Calculate the new freezing point of the solution.
The freezing point of pure water is \(0 \,^\circC\). The freezing point of the solution is the freezing point of the pure solvent minus the depression. \[ Freezing Point_{solution} = Freezing Point_{water} - \Delta T_f \] \[ Freezing Point_{solution} = 0 \,^\circC - 0.186 \,^\circC = -0.186 \,^\circC \] Quick Tip: For colligative properties, always check if the solute is an electrolyte or non-electrolyte to determine the van't Hoff factor (\(i\)). For non-electrolytes like urea, glucose, and sucrose, \(i=1\). For electrolytes like NaCl, \(i=2\).

Step 1: Recall the mechanism of ozone depletion.
The ozone layer is depleted by free radical catalysts. The most well-known are chlorine radicals (from CFCs) and nitric oxide radicals.

Step 2: Analyze the role of nitric oxide (NO).
Nitric oxide, often produced by the exhaust of supersonic jets in the stratosphere, acts as a catalyst in a cycle that converts ozone (\(O_3\)) to oxygen (\(O_2\)). The catalytic cycle is as follows: \[ NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g) \] \[ NO_2(g) + O(g) \rightarrow NO(g) + O_2(g) \]
The NO molecule is regenerated in the second step, allowing it to destroy thousands more ozone molecules. Quick Tip: Remember that ozone depletion involves {catalytic cycles}. The active species (like Cl\(\cdot\) or NO) is regenerated, which is why a small amount can cause significant damage to the ozone layer.

Step 1: Understand precipitation in coordination compounds.
When silver nitrate (AgNO\(_3\)) is added to a solution of a coordination complex, it only reacts with the halide ions that are outside the coordination sphere (the square brackets). These are known as counter-ions. Ions inside the coordination sphere are covalently bonded to the central metal atom and do not dissociate.

Step 2: Relate the amount of precipitate to the number of counter-ions.
The problem states that one mole of silver chloride (AgCl) is precipitated per mole of the complex. This means there must be exactly one chloride ion (\(Cl^-\)) available as a counter-ion in the formula.

Step 3: Examine the given formulas.
- (A) \([CoCl_2(NH_3)_4]Cl\): Has one \(Cl^-\) outside the brackets. Will precipitate 1 mole of AgCl.
- (B) \([CoCl(NH_3)_5]Cl_2\): Has two \(Cl^-\) outside the brackets. Will precipitate 2 moles of AgCl.
- (C) \([CoCl_3(NH_3)_3]\): Has zero \(Cl^-\) outside the brackets. Will precipitate 0 moles of AgCl.
- (D) \([Co(NH_3)_6]Cl_3\): Has three \(Cl^-\) outside the brackets. Will precipitate 3 moles of AgCl.

Based on this analysis, formula (A) is the correct answer. Quick Tip: Think of the square brackets in a coordination compound as a protective box. Reagents like AgNO\(_3\) can only react with the ions that are outside this box. The number of moles of AgCl formed directly tells you the number of Cl\(^-\) ions outside the brackets.

Step 1: Write the balanced chemical equation for the reaction.
The standard balanced equation for the oxidation of sulphur dioxide (\(SO_2\)) to sulphur trioxide (\(SO_3\)) is: \[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
This equation shows the reaction for 2 moles of \(SO_2\). The question asks for 4 moles of \(SO_2\), so we multiply the entire equation by 2: \[ 4SO_2(g) + 2O_2(g) \rightleftharpoons 4SO_3(g) \]

Step 2: Calculate \( \Delta n_g \).
The change in the number of moles of gas (\(\Delta n_g\)) is the difference between the total moles of gaseous products and the total moles of gaseous reactants. \[ \Delta n_g = (moles of gaseous products) - (moles of gaseous reactants) \]
From our balanced equation:
- Moles of gaseous products = 4 mol (from 4\(SO_3\))
- Moles of gaseous reactants = 4 mol (from 4\(SO_2\)) + 2 mol (from 2\(O_2\)) = 6 mol

Step 3: Substitute the values to find \( \Delta n_g \). \[ \Delta n_g = 4 - 6 = -2 \] Quick Tip: To find \( \Delta n_g \), always start with a balanced chemical equation for the specific quantities mentioned. Then, carefully sum the stoichiometric coefficients of all gaseous products and subtract the sum of the coefficients of all gaseous reactants.

Step 1: Understand the dimensionality of nanostructures.
Nanostructures are classified based on the number of dimensions that are not confined to the nanoscale range (typically 1-100 nm).
- 0D (Zero-dimensional): All three dimensions are at the nanoscale. Example: Nanoparticles (quantum dots).
- 1D (One-dimensional): Two dimensions are at the nanoscale, while the third is larger, creating an elongated structure. Examples: Nanotubes, nanorods, nanowires.
- 2D (Two-dimensional): One dimension is at the nanoscale, while the other two are larger, creating a sheet-like structure. Example: Nanofilms (graphene).
- 3D (Three-dimensional): None of the dimensions are confined to the nanoscale. These are bulk materials made of nanoscale components.

Step 2: Classify the options.
- (A) Nanoparticles are 0D.
- (B) Nanotubes are 1D.
- (C) Nanofilms are 2D.
- (D) Nanorods are 1D.
Both Nanotubes and Nanorods fit the description of a one-dimensional nanostructure. Quick Tip: Think of the "D" as the number of "free" or large dimensions. - 0D: A point (all small) \(\rightarrow\) Nanoparticles. - 1D: A line (long in one direction) \(\rightarrow\) Nanotubes, nanorods. - 2D: A plane (large in two directions) \(\rightarrow\) Nanofilms.

Step 1: State Ostwald's Dilution Law.
This law relates the dissociation constant (\(K_a\)) of a weak electrolyte, its degree of dissociation (\(\alpha\)), and its concentration (\(c\)).

Step 2: Derive the relationship.
Consider a weak monoprotic acid HA dissociating in solution: \[ HA \rightleftharpoons H^+ + A^- \]
Initial concentration: \(c\) \quad 0 \quad 0
Equilibrium concentration: \(c(1-\alpha)\) \quad \(c\alpha\) \quad \(c\alpha\)

The acid dissociation constant, \(K_a\), is given by: \[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha} \]
For a weak electrolyte, the degree of dissociation \(\alpha\) is very small, so we can approximate \(1-\alpha \approx 1\). \[ K_a \approx c\alpha^2 \]
Step 3: Rearrange the formula to solve for \(\alpha\). \[ \alpha^2 = \frac{K_a}{c} \quad \Rightarrow \quad \alpha = \sqrt{\frac{K_a}{c}} \]
This shows that the degree of dissociation is inversely proportional to the square root of the concentration. Quick Tip: Ostwald's Dilution Law is a cornerstone for understanding weak electrolytes. The key approximation is \(1-\alpha \approx 1\), which simplifies the math and leads directly to the relationship \( \alpha = \sqrt{K_a/c} \).

Step 1: Analyze the effect of substituents on the acidity of benzoic acid.
Acidity is determined by the stability of the conjugate base (carboxylate anion, -COO\(^-\)). Electron-withdrawing groups (EWGs) stabilize the anion and increase acidity. Electron-donating groups (EDGs) destabilize the anion and decrease acidity.

Step 2: Evaluate each compound.
- (C) Benzoic acid: This is our reference compound.
- (A) o-Hydroxybenzoic acid (Salicylic acid): The -OH group is an EWG via its inductive effect (-I) but an EDG via resonance (+R). However, at the ortho position, the conjugate base is significantly stabilized by intramolecular hydrogen bonding between the -OH proton and the -COO\(^-\) group. This is a powerful stabilizing effect known as the "ortho effect," making it much more acidic than benzoic acid.
- (B) o-Aminobenzoic acid: The -NH\(_2\) group is a strong EDG via resonance (+R), which destabilizes the conjugate base and decreases acidity.
- (D) p-Methoxybenzoic acid: The -OCH\(_3\) group at the para position is a strong EDG via resonance (+R), destabilizing the conjugate base and making it less acidic than benzoic acid.

Step 3: Compare the acidities.
Due to the strong stabilization from intramolecular hydrogen bonding, o-hydroxybenzoic acid is the most acidic compound among the choices. The general order is: (A) > (C) > (D) > (B). Quick Tip: When comparing substituted benzoic acids, look for the "ortho effect." Ortho substituents, especially those capable of hydrogen bonding (like -OH, -NH\(_2\)), can have a surprisingly large impact on acidity due to steric hindrance and intramolecular H-bonding that stabilizes the conjugate base.

Step 1: Define the terms.
- Molar conductivity (\(\Lambda_m\)): The conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.
- Conductivity (k or \(\kappa\)): The conductance of a solution of 1 cm length with a cross-sectional area of 1 cm\(^2\). Its unit is S cm\(^{-1}\).
- Concentration (c): The amount of electrolyte in moles per liter (mol L\(^{-1}\)).

Step 2: Relate the terms.
Molar conductivity is defined as the conductivity divided by the molar concentration: \[ \Lambda_m = \frac{k}{c} \]
However, the units must be consistent. Typically, \(k\) is in S cm\(^{-1}\) and \(c\) is in mol L\(^{-1}\). To make them compatible, we must convert the volume from Liters to cm\(^3\) (since 1 L = 1000 cm\(^3\)). \[ \Lambda_m (S cm^2 mol^{-1}) = \frac{k (S cm^{-1})}{c (mol L^{-1})} \times \frac{1000 (cm^3)}{1 (L)} \] \[ \Lambda_m = \frac{1000k}{c} \] Quick Tip: The "1000" in the molar conductivity formula is a unit conversion factor, not a fundamental constant. It's used specifically when concentration is in mol/L and conductivity is in S/cm. Always be mindful of the units you are given and the units you need.

Step 1: Define the classes of amines.
Amines are classified based on the number of alkyl or aryl groups directly bonded to the nitrogen atom.
- Primary (1\(^\circ\)) amine: One carbon group attached to nitrogen (general formula R-NH\(_2\)).
- Secondary (2\(^\circ\)) amine: Two carbon groups attached to nitrogen (general formula R\(_2\)-NH).
- Tertiary (3\(^\circ\)) amine: Three carbon groups attached to nitrogen (general formula R\(_3\)-N).

Step 2: Classify each option.
- (A) Cyclohexylamine: The nitrogen is attached to one cyclohexyl group. It is a primary amine (C\(_6\)H\(_{11}\)-NH\(_2\)).
- (B) Isopropylamine: The nitrogen is attached to one isopropyl group. It is a primary amine ((CH\(_3\))\(_2\)CH-NH\(_2\)).
- (C) Diphenylamine: The nitrogen is attached to two phenyl groups. It is a secondary amine ((C\(_6\)H\(_5\))\(_2\)NH).
- (D) N, N-Dimethylaniline: The nitrogen is attached to one phenyl group and two methyl groups, for a total of three carbon groups. It is a tertiary amine.

Therefore, diphenylamine is the secondary amine. Quick Tip: To quickly classify an amine, count the number of hydrogen atoms attached to the nitrogen. Two H's = Primary. One H = Secondary. Zero H's = Tertiary. (This trick works for simple, non-quaternary amines).