Step 1: Identify the Elements

The prompt states that both A²⁺ and B²⁺ ions have 4 unpaired electrons. Let's analyze the 3d configurations:

Step 2: Assign A and B based on Density

The prompt also states that element (A) is less dense than element (B).

Therefore, A = Chromium (Cr) and B = Iron (Fe).

Step 3: Match Uses

Comparing with the options, (c) is the correct match.

Step 1: Identify the Elements

Step 2: Analyze the Stability of Conversions

A conversion to a "lower energy" compound means forming a more stable product. We check the oxidation states.

Step 1: Identify Elements A, B, and C

Step 2: Order by Density

Density generally increases across a period in the d-block. Let's compare the densities of the identified elements.

The correct order of increasing density is:

Step 1: Identify Element B and the Sequence

Given: B³⁺ has the configuration [Ar] 3d³. To find the neutral atom B, we add back 3 electrons. They go into the 4s and 3d orbitals.

Neutral B configuration: [Ar] 4s² 3d⁴. This is the expected configuration for Chromium (Cr), Atomic Number 24 (although its actual configuration is anomalous: 4s¹3d⁵).

The four consecutive elements are:

Step 2: Evaluate Statement (c)

Statement (c) is about element C = Manganese (Mn). It describes losing "the first electron from the (d) sublevel". This typically refers to the oxidation process from a common ion, in this case Mn²⁺ → Mn³⁺.

The number of unpaired electrons decreases from 5 to 4. Since magnetic moment is directly related to the number of unpaired electrons, the magnetic moment decreases. This matches the statement.

Key Concept: Magnetic Moment of Iron

To increase the magnetic moment, we need to increase the number of unpaired electrons. This means oxidizing Fe²⁺ to Fe³⁺.

Analyzing the Options:

(a) FeCl₃ + NH₄OH → Fe(OH)₃ → Fe₂O₃: The iron starts as Fe³⁺ and ends as Fe³⁺. No change.

(b) Fe + H₂SO₄ → FeSO₄; then heat:

• Step 1: Fe + H₂SO₄(dil) → FeSO₄ + H₂. The compound formed is Iron(II) sulfate (Fe²⁺).
• Step 2: 2FeSO₄(s) --(heat)→ Fe₂O₃(s) + SO₂(g) + SO₃(g). The final product is Iron(III) oxide (Fe³⁺).

(c) Limonite (Fe₂O₃·nH₂O) heated: Starts as Fe³⁺ and ends as Fe³⁺. No change.

(d) Siderite (FeCO₃) heated in absence of air: FeCO₃ → FeO + CO₂. The iron starts as Fe²⁺ and ends as Fe²⁺. No change.

Step 1: Analyze the Clues

• The anion is X²⁻ (divalent). This eliminates Nitrite (NO₂⁻).
• The gas (Y) "oxidizes with normal oxidizing agents", which means Y is a reducing agent. This eliminates Carbonate (CO₂), as carbon is already in its highest oxidation state (+4).

Step 2: Evaluate Remaining Options

Sulphite (SO₃²⁻):

• Reaction: SO₃²⁻ + 2H⁺ → H₂O + SO₂↑
• Gas (Y) is SO₂.
• SO₂ is a known reducing agent and has a pungent odor.
• It famously reduces acidified potassium dichromate from orange (Cr₂O₇²⁻) to green (Cr³⁺). This matches the description perfectly.

Sulphide (S²⁻):

• Reaction: S²⁻ + 2H⁺ → H₂S↑
• Gas (Y) is H₂S.
• H₂S has a strong, unpleasant "rotten egg" smell, not "odorless" as described in option C.

Therefore, the only option that fits all criteria is B.

A good distinguishing test must give a clearly different result for each substance. Let's analyze the reagents:

Since hot concentrated sulfuric acid produces brown fumes with both nitrate and nitrite, it is not a preferred method for telling them apart.

The described reactions are characteristic tests for halide and nitrate ions using concentrated sulfuric acid (H₂SO₄).

Therefore, Acid (X) is concentrated H₂SO₄, and the salts contain iodide, bromide, and nitrate ions, matching option (D).

Step 1: Analyze the Diagram

Barium chloride (BaCl₂) reacts with aqueous solutions X and Y to form two different precipitates (A and B). The crucial clue is that both precipitates are soluble in dilute acids (like HCl).

Step 2: Evaluate Barium Precipitates

Step 3: Eliminate Options

• Options (a), (b), and (c) contain either Na₂SO₄ as a reactant (which would form insoluble BaSO₄) or list BaSO₄ as a product. Since the diagram specifically states both precipitates are soluble in dilute acid, any option involving sulfate is incorrect.
• Option (d) pairs Na₂CO₃ (forms BaCO₃, soluble) and Ba₃(PO₄)₂ (soluble). Both meet the condition perfectly.

Step 1: Analyze the Clues

• Clue 1: The acid derived from Salt A detects the cation of Salt B.
• Clue 2: The acid derived from Salt B detects the anion of Salt A.

Step 2: Test Option (c)

Let's assign A = KCl and B = PbSO₄.

Salt A is KCl.
The acid of Salt A is Hydrochloric Acid (HCl).
Does HCl detect the cation of Salt B (Pb²⁺)?
Yes! Dilute HCl is the group reagent for Analytical Group I cations (Ag⁺, Pb²⁺, Hg₂²⁺). It reacts to form a white precipitate: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Salt B is PbSO₄.
The acid of Salt B is Sulfuric Acid (H₂SO₄).
Does concentrated H₂SO₄ detect the anion of Salt A (Cl⁻)?
Yes! Concentrated H₂SO₄ is the group reagent for halides. It reacts with solid chlorides to evolve acidic HCl gas: 2KCl + H₂SO₄ → K₂SO₄ + 2HCl↑

Since both conditions hold perfectly, Option (c) is the correct answer.

Le Chatelier's Principle

When the partial pressure of nitrogen (N₂) is abruptly increased:

1. Immediate Effect: The concentration/pressure of N₂ undergoes a vertical "spike" at the moment of addition. (This eliminates graph 'c', which shows no spike).
2. Equilibrium Shift: The system shifts to the right (towards products) to consume the added N₂.
3. Gradual Changes:
   • N₂ is consumed, so its pressure curves back down (but ends up higher than its original state).
   • H₂ is also consumed as it reacts with the N₂, so its pressure curves downward.
   • NH₃ is produced, so its pressure curves upward.

Graph (d) perfectly illustrates this: a sharp vertical spike in N₂ followed by a gradual decrease, accompanied by a decrease in H₂ and an increase in NH₃.

Conditions for Equilibrium

For a chemical or physical equilibrium to exist, the process must be reversible and occur in a system where reactants and products cannot escape (a closed system, or involving species that don't evaporate/precipitate out of the medium entirely).

Decomposition of hydrogen peroxide is a highly spontaneous, one-way reaction. The produced oxygen gas escapes, making the process irreversible under normal conditions.

Relationship between pH, Acid Strength, and Ka

For acids of the same concentration:

• A lower pH means a higher concentration of hydronium ions [H⁺].
• A higher [H⁺] implies the acid dissociates more completely, meaning it is a stronger acid.
• Stronger acids have larger ionization constants (Ka).

Therefore, the order of decreasing Ka is A > C > B.

Step 1: Calculate the Equilibrium Constant (Kc)

Because the temperature is constant, Kc remains the same before and after adding substance A.

Using the first set of equilibrium concentrations:

Kc = (3² × 4.5³) / (1.5² × 2.3³) = (9 × 91.125) / (2.25 × 12.167) = 820.125 / 27.37575 ≈ 29.958

Step 2: Solve for the new [B]

Using the new equilibrium concentrations and the same Kc:

29.958 = (4² × 6³) / (2.5² × [B]³)

29.958 = (16 × 216) / (6.25 × [B]³)

29.958 = 3456 / (6.25 × [B]³)

[B]³ = 3456 / (6.25 × 29.958) = 3456 / 187.2375 ≈ 18.458

[B] = ∛18.458 ≈ 2.64 M

Step 1: Determine the Partial Pressures

The reaction is: COCl₂(g) ⇌ CO(g) + Cl₂(g)

Because the products CO and Cl₂ are produced in a 1:1 molar ratio from COCl₂, their partial pressures at equilibrium must be equal.

P(Cl₂) = 0.3 atm, therefore P(CO) = 0.3 atm.

Step 2: Find the Pressure of COCl₂

The total pressure is the sum of the partial pressures of all gases:

P(Total) = P(COCl₂) + P(CO) + P(Cl₂)

0.8 atm = P(COCl₂) + 0.3 atm + 0.3 atm

0.8 atm = P(COCl₂) + 0.6 atm

P(COCl₂) = 0.2 atm

Step 3: Calculate Kp

Kp = (0.3 × 0.3) / 0.2 = 0.09 / 0.2 = 0.45

Reasoning: The ionization constant (Ka) is a thermodynamic value that depends only on temperature. Diluting a weak acid changes its degree of ionization (α), hydronium concentration, and pH, but it does NOT change the Ka value. Therefore, Ka of X equals Ka of Y.

Step 1: Determine EMF.
The reaction shows Cu²⁺ is reduced (Cathode) and Cr is oxidized (Anode).
Reduction potential of Cu = +0.340 V. Oxidation potential of Cr = +0.740 V.
EMF = E°(oxidation) + E°(reduction) = 0.740 V + 0.340 V = +1.08 V.

Step 2: Determine Cell Type.
Because the EMF is positive, the reaction is spontaneous. A spontaneous electrochemical cell is a Galvanic cell.

Connecting vessel (1) [Zn/Zn²⁺] and (4) [Cu/Cu²⁺] creates a classic Daniell cell. Zinc is more active than Copper, so Zinc acts as the anode (oxidation) and Copper acts as the cathode (reduction). Electrons flow through the wire from Anode (Zn, vessel 1) to Cathode (Cu, vessel 4).

Correction Note: In the user-provided prompt 'Answer Key', the answer is marked as (C) for 19. Let's recalculate based on standard chemistry.

First cell: X is Anode. E_cell = E_ox(X) + E_red(H) => 0.23 = E_ox(X) + 0. So, Oxidation Potential of X = +0.23 V.

Second cell: Y is Cathode. E_cell = E_ox(H) + E_red(Y) => 0.80 = 0 + E_red(Y). So, Reduction Potential of Y = +0.80 V. (Oxidation Potential of Y = -0.80 V).

Comparing Oxidation Potentials: X (+0.23 V) > Y (-0.80 V).
Because X has a higher oxidation potential, X is the Anode and Y is the Cathode.

EMF = E_ox(Anode) + E_red(Cathode) = 0.23 V + 0.80 V = +1.03 V.

(Thus, statement (a) is chemically correct based on the provided text, despite what a raw key might say. X oxidizes, Y reduces.)

Understanding Protection Types

In standard curriculum terminology:

• Anodic Protection (Sacrificial Anode): Coating a metal with a more active metal. The coating has a higher oxidation potential.
• Cathodic Protection (Cathodic Coating): Coating a metal with a less active metal. The coating has a lower oxidation potential.

Ordering the Elements by Activity

Higher oxidation potential = More active.

W (+1.67) > X (+0.126) > Y (-0.401) > Z (-1.420)

Evaluating the Options

Reasoning: In an electrolytic cell, the Anode is the positive electrode where Oxidation (loss of electrons) takes place.

Molten potassium bromide (KBr) contains K⁺ and Br⁻ ions.

• At the Cathode (Reduction): K⁺ + e⁻ → K(l)
• At the Anode (Oxidation): 2Br⁻ → Br₂(g) + 2e⁻

In a Hydrogen-Oxygen fuel cell (alkaline type), the electrolyte is typically hot aqueous KOH. Let's look at the half-reactions:

Anode (Oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Here, OH⁻ ions are consumed.

Cathode (Reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Here, OH⁻ ions are formed.

Therefore, the hydroxide ions migrate from the cathode to the anode, being produced at the cathode (reduction site) and used up at the anode (oxidation site).

Step 1: Identify Compound X

Dry distillation of sodium acetate with soda lime produces Methane:

Chlorinating Methane by replacing three hydrogen atoms yields Chloroform (CHCl₃):

Properties of X: Chloroform is a substituted alkane. It was historically used as an anesthetic.

Step 2: Identify Compound Y

Dehydration of 2-chloroethanol (Cl-CH₂-CH₂-OH) at 180°C removes H and OH to form a double bond:

Properties of Y: The product is Vinyl Chloride (chloroethene). It is an alkene derivative used as the monomer to make PVC (Polyvinyl Chloride), which is used in water hoses and pipes. This makes option (d) true regarding compound Y,

Step 1: Determine the molecular formula.
Combustion of a hydrocarbon C_xH_y produces x moles of CO₂ and y/2 moles of H₂O.
If 5 moles of CO₂ are produced, x = 5 (Carbon atoms = 5).
If 5 moles of H₂O are produced, y/2 = 5 → y = 10 (Hydrogen atoms = 10).
The required molecular formula is C₅H₁₀.

Step 2: Check the options.

Chemical Analysis: The formula C₄H₈O corresponds to an aldehyde or a ketone.

• X oxidizes (turns dichromate green): X must be an aldehyde (e.g., butanal or 2-methylpropanal). Option C correctly identifies X as 2-methyl propanal (C₄H₈O).
• Y does not oxidize: Y must be a ketone (e.g., butanone). Option D suggests Y is 3-methyl-2-butanone, but that compound has 5 carbons (C₅H₁₀O), not 4, .

Step 1: Identify the Monomer

The repeating unit of the polymer is —CH₂—CH(C₂H₅)—. To find the monomer, we restore the double bond between the two carbon atoms of the backbone.

Step 2: Evaluate the Reactions of 1-Butene

1-Butene is an unsymmetrical alkene, so its addition reactions follow Markovnikov's Rule (the hydrogen adds to the carbon with more hydrogens in case of addition of asymmetric reagents only not all addition reactions).

Identify Compounds:

• A (C₂H₄O): Acetaldehyde (Ethanal)
• B (C₂H₆O): Ethanol (Ethyl alcohol)
• C (C₂H₄): Ethene (Ethylene)

Analyze Oxidation:
Oxidation of Acetaldehyde (A) yields Acetic acid (Ethanoic acid).
Oxidation of Ethanol (B) yields Acetaldehyde, then Acetic acid.
Acetic acid is a well-known raw material used in the chemical industry for the manufacture of dyes, plastics, and insecticides.

The compound with the formula C₇H₇NO is Benzamide (C₆H₅CONH₂). Count the atoms: C: 6+1=7, H: 5+2=7, N: 1, O: 1.

Amides not ester are prepared by the ammonolysis of esters (reacting an ester with ammonia).

Reacting Methyl benzoate (Option A) with ammonia perfectly yields Benzamide and Methanol.

Match formulas to functional groups:

• X (CH₂O): Formaldehyde (Methanal). This is an Aldehyde. (Ketones require at least 3 carbons).
• Y (C₂H₆O): Ethanol (CH₃CH₂OH) or Dimethyl ether. Ethanol is a Primary monohydric alcohol.
• Z (C₂H₆O₂): Ethylene glycol (HO-CH₂-CH₂-OH). This is a Primary dihydric alcohol.

Option B aligns perfectly with these structures.

Step 1: Identify C
Compound C does not react with NaOH, but it reacts with A to form a flavor (an ester). This means C is an alcohol. C₂H₆O is Ethanol.

Step 2: Identify A and B based on NaOH consumption

• A and B must be acids to react with NaOH.
• C₃H₆O₃ (Lactic acid): Contains one -COOH group and one aliphatic -OH group. Aliphatic -OH does not react with NaOH. So, it consumes 1 mole of NaOH.
• C₇H₆O₃ (Salicylic acid): Contains one -COOH group and one phenolic -OH group. Phenols DO react with NaOH. So, it consumes 2 moles of NaOH.

Since B consumes double the quantity of A, B is Salicylic acid (C₇H₆O₃) and A is Lactic acid (C₃H₆O₃). Heating A (Lactic acid) with C (Ethanol) produces Ethyl lactate, an ester used as a flavoring agent.

The boiling point of organic compounds with hydroxyl (-OH) groups is primarily determined by the extent of hydrogen bonding. More -OH groups mean stronger intermolecular forces.

Therefore, the correct decreasing order is A > B > C.

The compound C₆H₅OOCCH₃ is an ester. Specifically, it is formed from the acetate (ethanoate) radical CH₃COO- and the phenyl radical -C₆H₅. Its common name is phenyl acetate.

Acidic hydrolysis of an ester breaks it back down into its constituent carboxylic acid and alcohol/phenol:

The IUPAC name for CH₃COOH is Ethanoic acid. The IUPAC accepted name for C₆H₅OH is Phenol. (While "acetic acid" and "hydroxybenzene" are valid names, "ethanoic acid" and "phenol" perfectly fit standard IUPAC multiple-choice criteria here).

Step 1: Identify X and Y.
1. Identify elements $X$ and $Y$: * The question states that $X$ and $Y$ are two consecutive transition elements in the first transition series, and each has a compound in its highest oxidation state that acts as an oxidizing agent. * Chromium ($\text{Cr}$, $Z = 24$): Its highest oxidation state is $+6$, found in potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), which is a well-known strong oxidizing agent. * Manganese ($\text{Mn}$, $Z = 25$): Its highest oxidation state is $+7$, found in potassium permanganate ($\text{KMnO}_4$), which is also a powerful oxidizing agent. * Therefore, $X$ and $Y$ are $\text{Cr}$ and $\text{Mn}$.

Step 2: Evaluate the uses.
If we let Y = Cr(V2O5) and Mn = Mn (or vice versa):

• (a) Iron (Fe) is the catalyst for ammonia (Haber process), not Cr or Mn.
• (b) Vanadium ($\text{V}$), not Chromium ($\text{Cr}$), is used in car springs., .
• Alloys of Nickel with Chromium or steel exhibit acid resistance, which does not match the description of the element following $\text{Cr}$ (Manganese).
• The elements between which $\text{Cr}$ ($Z=24$) and $\text{Mn}$ ($Z=25$) are located are Vanadium ($\text{V}$, $Z=23$) and Iron ($\text{Fe}$, $Z=26$). * An alloy of Vanadium ($\text{V}$) and Iron ($\text{Fe}$) with carbon forms vanadium steel, which is characterized by high hardness and high elasticity (used in making car springs).

1. Identify the precipitates: AgNO₃ forms Ag₃PO₄ (yellow, soluble in NH₄OH) and AgI (yellow, insoluble in NH₄OH).

2. Mass of AgI: Since excess NH₄OH dissolves the Ag₃PO₄, the remaining 8.5g is entirely AgI.
Moles of AgI = 8.5g / (108+127)g/mol = 8.5 / 235 = 0.03617 mol.

3. Find KI mass in mixture: Moles KI = Moles AgI = 0.03617 mol.
Mass KI = 0.03617 × (39+127) = 0.03617 × 166 = 6.00 g.

4. Find Na₃PO₄ mass: Total mixture = 12g. Mass Na₃PO₄ = 12g - 6g = 6.00 g.
Moles Na₃PO₄ = 6.00 / (23×3 + 31 + 16×4) = 6.00 / 164 = 0.03658 mol.

5. Find Ag₃PO₄ mass: Moles Ag₃PO₄ = Moles Na₃PO₄ = 0.03658 mol.
Mass Ag₃PO₄ = 0.03658 × (108×3 + 31 + 16×4) = 0.03658 × 419 = 15.32 g.

6. Total initial yellow precipitate (X): Mass AgI + Mass Ag₃PO₄ = 8.5g + 15.32g = 23.82 g.

Step 1: Write the balanced chemical equation.

From the equation, 1 mole of acid reacts with 2 moles of base (nₐ = 1, n_b = 2).

Step 2: Use the titration formula.

(Mₐ × 32) / 1 = (M_b × 16) / 2

32 Mₐ = 8 M_b

Mₐ = (8 / 32) M_b = ¼ M_b

Therefore, the concentration of the sulphuric acid is a quarter of the concentration of the potassium hydroxide.

Step 1: Convert solubility to Molarity (mol/L).
Solubility given = 4.4 × 10⁻⁵ mol / 100 mL.
To find mol / 1000 mL (1 Liter), multiply by 10:
Molar Solubility (s) = 4.4 × 10⁻⁴ mol/L.

Step 2: Write the dissociation equation and Ksp expression.
MgCO₃(s) ⇌ Mg²⁺(aq) + CO₃²⁻(aq)
Ksp = [Mg²⁺][CO₃²⁻] = (s)(s) = s².

Step 3: Calculate Ksp.
Ksp = (4.4 × 10⁻⁴)² = 19.36 × 10⁻⁸ = 1.936 × 10⁻⁷.

Step 1: ICE Table (Initial, Change, Equilibrium)

The graph must show reactant [A] starting at 0.8 and dropping to level off at 0.4. Product [B] must start at 0 and rise to level off at 0.8. (Graph B illustrates this perfectly).

Step 2: Calculate Kc

Kc = (0.8)² / (0.4) = 0.64 / 0.4 = 1.6.

Step 1: Electricity for Gold
Reaction: Au³⁺ + 3e⁻ → Au
To precipitate 1 mole of Au, we need 3 moles of electrons (3 Faradays).

Step 2: Determine valence of unknown metal
We have 3 Faradays of electricity available. We want to know which ion will yield 1.5 moles of precipitate with these 3 Faradays.
Moles of metal = (Faradays) / (Valence charge z)
1.5 = 3 / z
z = 3 / 1.5 = 2.

The unknown ion must have a +2 charge. Looking at the options, only Mg²⁺ is divalent (+2).

From the balanced half-reaction, producing 1 mole of MnO₄⁻ requires 5 moles of electrons (5 Faradays).

To produce 0.2 moles of MnO₄⁻, the moles of electrons required = 0.2 × 5 = 1 mole of electrons (1 Faraday).

We know that 1 Faraday is equal to 96500 Coulombs (C). Therefore, the required electricity is 96500 C.

The formula C₃H₈O represents saturated alcohols and ethers. Let's analyze the clues:

• Compound X: Oxidizes to an acid (reacts with Na₂CO₃). Therefore, X is a primary alcohol. Structure: CH₃-CH₂-CH₂OH (Propan-1-ol). It has 1 methyl group.
• Compound Y: Oxidizes to a compound that isn't an acid (it's a ketone). Therefore, Y is a secondary alcohol. Structure: CH₃-CH(OH)-CH₃ (Propan-2-ol). It has 2 methyl groups.
• Compound Z: Does not oxidize. Therefore, Z is an ether. Structure: CH₃-O-CH₂-CH₃ (Ethyl methyl ether). It has 2 methyl groups.

This perfectly aligns with Option D: 1, 2, 2.

A "longer and weaker" O-H bond means the hydrogen is more easily lost as an H⁺ ion, indicating stronger acidity. In organic chemistry, phenols are more acidic than aliphatic alcohols because the benzene ring withdraws electron density, weakening the O-H bond.

Therefore, (A) is a Phenol and (B) is an Alcohol.

• Phenols (A) are acidic enough to react with strong bases like NaOH to form sodium phenoxide.
• Alcohols (B) are practically neutral and do not react with NaOH.

Step 1: Trace the reactions.

• X: Dry distillation of Sodium Decanoate (C₉H₁₉COONa) yields an alkane with one less carbon: Nonane (C₉H₂₀).
• Y & Z: Thermal cracking of Nonane yields a smaller alkane and an alkene. Z is the "simplest alkene" (Ethene, C₂H₄). So Y is Heptane (C₇H₁₆).
• W: Catalytic reforming of Heptane yields Toluene (methylbenzene, C₆H₅CH₃), an aromatic hydrocarbon.
• M: Hydrogenation of Toluene (adding H₂ to break the aromatic ring) yields Methylcyclohexane.

Step 2: Evaluate compound properties.

M (Methylcyclohexane) is a saturated cyclic hydrocarbon (but NOT cyclohexane exactly, making B incorrect).
W (Toluene) oxidizes with KMnO₄ to form Benzoic acid (an aromatic acid).

Option A perfectly describes the families of these compounds.

• Acid A: Alkaline hydrolysis of 1,1,1-trichloroethane (CH₃CCl₃ + 3NaOH) temporarily forms a tri-alcohol CH₃C(OH)₃, which is highly unstable. It instantly loses a water molecule to become Acetic acid (CH₃COOH). Formula: C₂H₄O₂.
• Acid B: Fermentation of milk sugars (lactose) yields Lactic acid. Formula: C₃H₆O₃.
• Acid C: The substance used for rheumatic pain is Methyl salicylate (oil of wintergreen). It is formed by reacting Methanol with Salicylic acid. Formula: C₇H₆O₃.

Salicylic Acid (X): Contains 1 Carboxyl group (-COOH) and 1 Phenolic group (-OH).
- NaOH reacts with BOTH (acids and phenols). Needs 2 moles.
- Na metal reacts with BOTH. Needs 2 moles.

Lactic Acid (Y): Contains 1 Carboxyl group (-COOH) and 1 Aliphatic Alcohol group (-OH).
- NaOH reacts ONLY with the acid group. Needs 1 mole.
- Na metal reacts with BOTH acid and alcohol groups. Needs 2 moles.

Chemistry Breakdown for 46:
Sucrose (A) hydrolysis yields Glucose and Fructose (isomers). Fermentation of Glucose (B) yields Ethanol(D).
Ethanol (D) + H₂SO₄ at 180°C → Ethene (Y).
Ethanol (D) + H₂SO₄ at 140°C → Diethyl ether (X).