(i) Substance 'X' is Calcium Carbonate.
Chemical formula: CaCO₃

(ii)
Reaction 1 – CaCO₃ is insoluble in water (no equation needed, it is a physical property).

Reaction 2 – CaCO₃ reacts with dilute HCl to produce CO₂ gas:

$$\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$$

Reaction 3 – CO₂ turns lime water milky:

$$\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$$

(White precipitate of CaCO₃ makes lime water milky.)

Source: Chapter 2, Acids, Bases and Salts; Chapter 1, Section 1.2.1

---

• The clues — building material, insoluble in water, reacts with acid to give CO₂ — all point to CaCO₃ (limestone/marble/chalk, all used in construction).
• Examiners expect both equations: the acid–carbonate reaction AND the lime water reaction. Missing either loses marks.
• Write state symbols — (s), (aq), (g), (l) — as CBSE awards a mark for balanced equations with correct states.
• The lime-water reaction is the same as equation 1.14 from the textbook (reverse context), so knowing it from Chapter 1 is helpful here.

(d) Assertion (A) is false, but Reason (R) is true.

The correct practice is to add acid to water (not water to acid). So Assertion (A) is false. The Reason (R) is true — dissolving acid in water is highly exothermic.

The textbook (Chapter 2, Activity 2.10) clearly states: "The acid must always be added slowly to water with constant stirring." The Assertion incorrectly reverses this — it says "add water to acid," which is the dangerous practice. The Reason is correctly stated (the process is highly exothermic). Hence, A is false but R is true → option (d).

(b) 'I' is a displacement reaction and 'II' is a double displacement reaction.

In equation I, Mg displaces H from HCl (single displacement). In equation II, Na⁺ and H⁺ exchange partners — a double displacement (neutralisation) reaction.

• Displacement reaction: One element displaces another from a compound. Here, Mg displaces H₂ from HCl.
• Double displacement reaction: Ions of two compounds exchange places. In NaOH + HCl, Na⁺ pairs with Cl⁻ and H⁺ pairs with OH⁻ to form water — classic neutralisation = double displacement.
• Common mistake: Students label II as "neutralisation only" and forget neutralisation is a type of double displacement reaction.

(a) X = Na₂CO₃, Y = NH₄Cl

Phenolphthalein turns pink in basic solutions. Na₂CO₃ is a salt of strong base and weak acid, so its solution is basic (turns pink). NH₄Cl is a salt of strong acid and weak base, so its solution is acidic/neutral (no colour change).

Source: Chapter 2, Section 2.1.1

Phenolphthalein is colourless in acidic/neutral solutions and turns pink in basic solutions. Na₂CO₃ undergoes hydrolysis to give a basic solution; NH₄Cl gives an acidic solution. Options (b), (c), (d) are wrong because Na₂SO₄ and NaNO₃ give neutral solutions, and NaHCO₃ also gives a mildly basic solution — but the only pair where X is clearly basic (turns pink) and Y is acidic/neutral (no change) is option (a). Examiners expect you to recall the nature of salt solutions based on the strength of parent acid and base.

(c) acidic | strong acid and weak base | less than 7

Ammonium chloride is a salt of HCl (strong acid) and NH₄OH (weak base). Its solution is acidic, turning universal indicator pink, with pH less than 7.

Source: Chapter 2, Section 2.3

• Ammonium chloride (NH₄Cl) is formed from HCl (strong acid) + NH₄OH (weak base) → acidic salt.
• Salts of strong acid + weak base give acidic solutions (pH < 7).
• Pink colour on universal indicator confirms acidic nature (pH slightly below 7).
• Option (a) is wrong because it pairs weak acid + strong base (that gives basic salt). Options (b) and (d) are eliminated as they suggest basic/neutral nature. Only option (c) is correct.

The path of a beam of light passing through a colloidal solution is visible due to the Tyndall effect. The colloidal particles are relatively larger in size than the particles of a true solution. When light strikes these colloidal particles, they scatter light in all directions. This scattered light reaches our eyes, making the path of the beam visible. This phenomenon is called the Tyndall effect.

Source: Chapter 10, Section 10.6 – Scattering of Light / 10.6.1 Tyndall Effect

• The key term examiners look for is Tyndall effect — always name it.
• The core reason is scattering of light by colloidal particles (which are larger than particles in a true solution).
• Contrast with a true solution (path not visible) to strengthen the answer — it shows conceptual clarity.
• This is typically a 2–3 mark question; the answer above covers all required points concisely.

(a)
(i) During indigestion (excess acid in stomach), we take an antacid such as milk of magnesia (magnesium hydroxide) or sodium bicarbonate to neutralise the excess acid and raise the pH.

(ii) When stung by a honey bee (which injects formic acid/acidic venom) or nettle leaves, we apply a mild base such as baking soda (sodium bicarbonate) solution on the affected area to neutralise the acid and relieve the pain.

(b) When fresh milk (pH 6) changes to curd, its pH will decrease (become more acidic). This is because bacteria in milk convert lactose into lactic acid during fermentation, increasing the acidity of the curd.

---

• Part (a) tests the concept of neutralisation: excess acid → use a base; acidic sting → apply a base.
• Honey bee sting is acidic; wasp sting is alkaline (use acid to counteract). Know the difference.
• Part (b): curd formation = lactic acid production → lower pH. Examiners expect both the direction of change AND the reason (lactic acid formation).

(i) Compound X is Plaster of Paris.
Chemical formula: $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$

(ii) Baking soda contains only sodium hydrogen carbonate ($\text{NaHCO}_3$), whereas baking powder contains sodium hydrogen carbonate along with a mild edible acid (such as tartaric acid).

• For (i), Plaster of Paris is made from gypsum ($\text{CaSO}_4 \cdot 2\text{H}_2\text{O}$) by heating; its formula with the ½ water of crystallisation must be written correctly.
• For (ii), the key distinction is that baking powder = baking soda + mild acid (+ starch). Examiners expect you to name the extra component; just saying "it has more ingredients" is insufficient.

(c) Na₂CO₃.10H₂O

Washing soda is sodium carbonate decahydrate with the formula Na₂CO₃.10H₂O.

Washing soda = Na₂CO₃.10H₂O. NaHCO₃ is baking soda, NaOH is caustic soda, and Na₂CO₃.5H₂O is not the correct hydrated form. Examiners expect you to know the exact formula including the water of crystallisation (10H₂O).

(b) Milk of magnesia, Blue, 10, Basic.

Milk of magnesia (Mg(OH)₂) is a base with pH = 10, so it turns pH paper blue and is basic in nature. This is the only fully correct row.

Source: Chapter 2, Section 2.3

• Option (a) is wrong: lemon juice is acidic, not basic.
• Option (c) is wrong: gastric juice has pH ≈ 2, not 6 (it is strongly acidic, turning pH paper red/orange).
• Option (d) is wrong: pure water turns pH paper green, not yellow; pH = 7 (neutral) is correct but the colour is wrong.
• Option (b) is entirely correct — milk of magnesia is a base, pH 10, blue on pH paper.

Process: Chlor-Alkali Process

When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to give sodium hydroxide. This process is called the chlor-alkali process.

$$2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{Cl}_2\text{(g)} + \text{H}_2\text{(g)}$$

Gas X — Chlorine (Cl₂): obtained at the anode.

Gas Y — Hydrogen (H₂): obtained at the cathode.

Compound Z — Bleaching Powder [Ca(ClO)₂]

Chlorine gas (X) reacts with dry slaked lime [Ca(OH)₂] to produce bleaching powder. It is widely used in water treatment plants (to kill germs) and in the textile industry (for bleaching).

Chemical equation:

$$2\text{Ca(OH)}_2 + 2\text{Cl}_2 \rightarrow \text{Ca(ClO)}_2 + \text{CaCl}_2 + 2\text{H}_2\text{O}$$

Source: Chapter 2, Section 2.4.3

---

• The examiner expects the name and brief definition of the chlor-alkali process, the balanced equation, correct identification of both gases with their respective electrodes, the name of compound Z (bleaching powder), its two uses, and its formation equation — all five elements map to the 5 marks.
• A common mistake is reversing the electrodes: Cl₂ → anode, H₂ → cathode.
• Write the bleaching powder equation exactly as given in the textbook; the examiner checks it carefully.
• "Dry calcium hydroxide" = dry slaked lime = Ca(OH)₂ — use this term in your answer.

(i)

| Solution | pH | Type |
|---|---|---|
| B | 1 | (1) Strongly acidic |
| C | 13 | (2) Strongly alkaline |
| A | 4 | (3) Weakly acidic |
| D | 7 | (4) Neutral |
| E | 10 | (5) Weakly alkaline |

Increasing order of H⁺ ion concentration (lower pH → higher H⁺):

$$C < E < D < A < B$$

(pH 13 < 10 < 7 < 4 < 1)

---

(ii)

(1) Acidic Salt:
Name: Sodium bisulphate (Sodium hydrogen sulphate)
Formula: NaHSO₄
Parent acid: Sulphuric acid (H₂SO₄) — strong acid, partially neutralised
Parent base: Sodium hydroxide (NaOH)

(2) Basic Salt:
Name: Basic copper chloride (Copper hydroxychloride)
Formula: Cu(OH)Cl
Parent acid: Hydrochloric acid (HCl)
Parent base: Copper hydroxide [Cu(OH)₂] — strong base, partially neutralised

---

• For (i), remember: lower pH = more acidic = higher H⁺ concentration. pH 7 is neutral; below 7 is acidic; above 7 is alkaline. "Strongly" means close to 1 or 14; "weakly" means closer to 7.
• For the increasing H⁺ order, arrange from highest pH to lowest pH (C→E→D→A→B).
• For (ii), an acidic salt is formed when a strong acid partially replaces H⁺ with a metal — it still has replaceable hydrogen. A basic salt is formed when a strong base is only partially neutralised — it still contains OH⁻ groups. Examiners expect the formula, name, and parent acid/base for each.

Answer: A — Washing Soda

Washing Soda (Na₂CO₃·10H₂O) is used in the glass, soap, and paper industries, as well as for softening hard water.

This is from the Acids, Bases and Salts chapter (Chapter 2). Examiners expect you to recall the uses of washing soda specifically — glass, soap, paper industries, and water softening. Do not confuse it with Baking Soda (NaHCO₃), which is used in cooking and as an antacid. The source passages provided are from Chapter 4 and do not cover this topic, but the correct answer is firmly from the CBSE syllabus Chapter 2.

Option B — Turmeric and Litmus

Turmeric and litmus are natural indicators as both are obtained from natural sources (turmeric plant and lichen respectively), unlike methyl orange and phenolphthalein, which are synthetic indicators.

The key distinction is natural vs. synthetic indicators. Litmus (from lichen) and turmeric (from the turmeric plant) are natural indicators. Methyl orange and phenolphthalein are synthetic (man-made) indicators. The textbook (Ch. 2) lists litmus as a natural indicator and uses turmeric as an example of a natural substance that can indicate acids/bases. Eliminate options containing only synthetic indicators.

(a) Sodium chloride (NaCl) is formed from:

• Acid: Hydrochloric acid (HCl)
• Base: Sodium hydroxide (NaOH)

(b) In Calcium sulphate (CaSO₄):

• Cation: Ca²⁺ (Calcium ion)
• Anion: SO₄²⁻ (Sulphate ion)

(c) Salts are classified into families based on the cations and anions present in them. Sodium chloride (NaCl) contains the Na⁺ cation and Cl⁻ anion. Washing soda (Na₂CO₃) also contains the Na⁺ cation. Since both salts share the same cation (Na⁺), they belong to the sodium family of salts. Hence, the statement is justified.

Source: Chapter 2 – Acids, Bases and Salts

---

• (a): Always identify the parent acid (provides anion) and parent base (provides cation) of a salt. NaCl → HCl + NaOH.
• (b): Dissociate the formula: CaSO₄ → Ca²⁺ + SO₄²⁻. Cation is the positive ion, anion is the negative ion.
• (c): The key phrase from the passage is "family of salts is classified on the basis of cations and anions." Use this directly to justify: both NaCl and Na₂CO₃ share Na⁺, so same family. 2-mark justification needs both the rule AND its application.

(C) tartaric acid

Tamarind contains tartaric acid, which is acidic in nature and turns blue litmus red.

This is a direct factual question. Tamarind's sour taste is due to tartaric acid. Remember: acids turn blue litmus red. The other options — methanoic acid (ants/nettles), acetic acid (vinegar), oxalic acid (tomatoes/spinach) — are acids found in different sources.

(i) The tomato juice with pH 4.6 will taste sour. Since its pH is less than 7, it is acidic in nature, and acids taste sour.

(ii) A strong acid completely dissociates (ionises) in aqueous solution to produce a large number of H⁺ ions. A weak base partially dissociates in aqueous solution, producing a small number of OH⁻ ions.

(iii) Acid rain lowers the pH of river/lake water. Aquatic animals are adapted to survive within a specific pH range. When pH falls below this range due to acid rain, their survival becomes difficult and they may die.

---

• (i) Any pH < 7 = acidic = sour taste. Direct and simple — state it clearly.
• (ii) Key contrast: "complete ionisation" (strong acid) vs "partial ionisation" (weak base). Mention the ions produced (H⁺ and OH⁻ respectively).
• (iii) The logic chain is: acid rain → lower pH of water → outside tolerable range for aquatic life → difficult survival. Follow this chain for full marks.

(d) Na₂CO₃

Na₂CO₃·10H₂O (washing soda) has 10 water molecules of crystallisation, which is the maximum among the given compounds (FeSO₄·7H₂O, CuSO₄·5H₂O, CaSO₄·2H₂O).

Source: Chapter 2, Section 2.4.4

---

The water of crystallisation for each compound:

• FeSO₄·7H₂O → 7 molecules
• CuSO₄·5H₂O → 5 molecules
• CaSO₄·2H₂O → 2 molecules
• Na₂CO₃·10H₂O → 10 molecules ✓

The textbook explicitly states Na₂CO₃·10H₂O (washing soda). Note: FeSO₄·7H₂O is not mentioned in the passage but is standard NCERT knowledge; Na₂CO₃ with 10 water molecules is the clear winner. Always remember the formula of washing soda for board exams.

(i) Acid + Metal → Salt + Hydrogen gas

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$

Main product: Salt (zinc sulphate)

(ii) Acid + Base → Salt + Water (Neutralisation reaction)

$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

Main product: Salt (sodium chloride)

(iii) Acid + Carbonate → Salt + Carbon dioxide + Water

$$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2\uparrow + \text{H}_2\text{O}$$

Main product: Salt (sodium chloride)

Source: Chapter 2, Acids, Bases and Salts

---

• Each sub-part carries 1 mark: ½ mark for the balanced equation + ½ mark for naming the main product is the typical examiner split.
• The common product in all three reactions is a salt — examiners expect you to state this explicitly.
• Use a standard example (Zn/H₂SO₄, NaOH/HCl, Na₂CO₃/HCl) — these are directly from the textbook and are safe choices.
• Ensure arrows for gas evolution (↑) are shown where relevant; it demonstrates awareness of physical states.

(i) As an antacid:
Baking soda (NaHCO₃) is a mild base (alkaline in nature). It neutralises excess hydrochloric acid present in the stomach, relieving indigestion.
$$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$

(ii) As a constituent of baking powder:
On heating, NaHCO₃ decomposes to release CO₂ gas, which makes the dough soft and fluffy (acts as a leavening agent).
$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

(iii) In soda-acid fire extinguishers:
When NaHCO₃ reacts with sulphuric acid, CO₂ is produced in large quantities. CO₂ being heavier than air cuts off the oxygen supply and extinguishes the fire.
$$2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{CO}_2\uparrow$$

Source: Chapter 2, Acids, Bases and Salts

---

• Each part carries 1 mark. Examiner looks for: the chemical property (basic/alkaline nature → neutralisation; decomposition on heating; reaction with acid to give CO₂) AND the reason it is useful.
• Including balanced equations earns full marks and demonstrates the property clearly — always write them if you remember them.
• Key terms to use: neutralisation, decomposition, CO₂ released, alkaline nature.

(B) (ii) and (iii)

Washing soda (Na₂CO₃·10H₂O) and Plaster of Paris (CaSO₄·½H₂O) both contain water of crystallisation. Bleaching powder and baking soda do not contain water of crystallisation.

• Water of crystallisation = fixed number of water molecules in one formula unit of a salt.
• Washing soda: Na₂CO₃·10H₂O ✓
• Plaster of Paris: CaSO₄·½H₂O ✓
• Baking soda (NaHCO₃) and Bleaching powder (CaOCl₂) have no water of crystallisation.
• Examiners expect you to recall the chemical formulae of these common compounds to answer such questions correctly.

(D) Tartaric acid

Tamarind contains tartaric acid, which is acidic in nature and turns blue litmus red.

Tamarind is a natural source of tartaric acid. The question tests knowledge of naturally occurring acids. Acetic acid is in vinegar, methanoic acid (formic acid) is in ant stings, and oxalic acid is in tomatoes — none are the primary acid in tamarind. Tartaric acid is the correct answer and is a standard fact expected in CBSE exams.

(D) B < A < C

Ammonium chloride (strong acid + weak base → acidic, pH < 7), potassium nitrate (strong acid + strong base → neutral, pH = 7), sodium carbonate (strong base + weak acid → basic, pH > 7).

From section 2.4.2: salts of strong acid + weak base are acidic (pH < 7); strong acid + strong base are neutral (pH = 7); strong base + weak acid are basic (pH > 7). NH₄Cl = HCl + NH₄OH (weak base) → acidic; KNO₃ = HNO₃ + KOH (both strong) → neutral; Na₂CO₃ = NaOH + H₂CO₃ (weak acid) → basic. So ascending pH order: B < A < C.

Chemical formula of washing soda: Na₂CO₃.10H₂O

Preparation from baking soda:
Baking soda (NaHCO₃) is heated to obtain sodium carbonate (Na₂CO₃). Recrystallisation of sodium carbonate gives washing soda.

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$$

$$\text{Na}_2\text{CO}_3 + 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3.10\text{H}_2\text{O}$$

Two uses of washing soda:

1. It is used in glass, soap and paper industries.
2. It is used for removing permanent hardness of water.

Source: Chapter 2, Section 2.4.3

---

• The examiner expects the formula with water of crystallisation (10H₂O) — writing just Na₂CO₃ loses a mark.
• Both steps must be shown: heating NaHCO₃ → Na₂CO₃, then recrystallisation → washing soda. Equations add scoring value.
• For uses, pick any two from the textbook list. "Cleaning agent" and "removing hardness of water" are the most commonly expected answers.

The correct answer is C — they have a fixed number of molecules of water of crystallisation in one formula unit of these salts.

"Water of crystallisation" means the fixed number of water molecules chemically bonded in one formula unit of a salt crystal (e.g., Na₂CO₃·10H₂O, CuSO₄·5H₂O, CaSO₄·2H₂O, FeSO₄·7H₂O). This is the one property common to all four salts. They are not all coloured (washing soda is white), nor are they all acidic or basic — they are salts, which can be neutral, acidic, or basic depending on the parent acid/base.

(i) Common salt (NaCl) is obtained from hydrochloric acid (HCl) (strong acid) and sodium hydroxide, NaOH (strong base).

(ii) Sodium chloride is neutral in nature. It is formed by the neutralisation of a strong acid (HCl) and a strong base (NaOH), so its aqueous solution has pH = 7.

(iii) When electric current is passed through brine (aqueous NaCl), the process is called chlor-alkali process. The products obtained are:

| Product | Place of collection |
|---|---|
| Chlorine gas (Cl₂) | Anode |
| Hydrogen gas (H₂) | Cathode |
| Sodium hydroxide (NaOH) | Near cathode (in solution) |

Source: Acids, Bases and Salts, Chapter 2 (NCERT Class 10 Science)

---

• (i) Examiners expect both the name and formula of the acid and base.
• (ii) The key reasoning is "strong acid + strong base → neutral salt." Always state pH = 7.
• (iii) This is a standard 2-mark factual question. Naming all three products AND their correct positions (anode/cathode) is essential to score full marks. Missing the location loses a mark.

(d) Assertion (A) is false, but Reason (R) is true.

The correct procedure is to add acid slowly to water (not water to acid) with constant stirring, as dissolving acid in water is highly exothermic. Thus, Assertion (A) is false. Concentrated nitric acid is indeed soluble in water, so Reason (R) is true.

Source: Chapter 2, Section 2.2.1

---

The key fact from the textbook is: "The acid must always be added slowly to water with constant stirring" — never water to acid. So the Assertion reverses this and is false. The Reason states nitric acid is soluble in water, which is true (all common mineral acids are water-soluble). Since A is false but R is true, the answer is (d). Watch out — examiners frequently test this dilution rule as it is a safety precaution explicitly highlighted in the textbook.

(a) Calcium chloride

Calcium chloride is used as a drying agent to absorb moisture and prepare dry HCl gas, as it does not react with HCl.

The other options (CaO, Ca(OH)₂, CaCO₃) are basic substances that would react with acidic HCl gas, not just dry it. Only anhydrous CaCl₂ acts as a neutral drying agent suitable for HCl. Note: The source passages do not cover this topic; the answer is based on standard Class 10 Chemistry knowledge of drying agents.

When we eat sugary or starchy food, bacteria present in the mouth break down (degrade) the sugar and food particles remaining in the mouth. This process produces acids, which lower the pH of the mouth below 5.5.

At this low pH, the tooth enamel (made of calcium hydroxyapatite) gets corroded, leading to tooth decay.

Source: Chapter 2, Section 2.3.1 — pH change as the cause of tooth decay

---

• The key cause is bacterial degradation of sugar → acid production.
• Mention that pH drops below 5.5 (the threshold given in the textbook).
• Examiners expect: (1) bacteria in mouth, (2) degradation of sugar/food particles, (3) acid produced → pH falls below 5.5. All three points together make a complete answer for 2 marks.
• Do not confuse this with stomach acidity — this question is specifically about the mouth.

(i) Solution M gives red colour with universal indicator, indicating it is acidic (pH < 7). Therefore, solution M has more hydrogen ion (H⁺) concentration, since higher H⁺ concentration corresponds to lower pH and acidic nature.

(ii) When M and N are mixed and the result turns green, the pH is 7 (neutral). The salt formed is a neutral salt, formed by the reaction of a strong acid and a strong base. Salts of strong acid and strong base have pH = 7.

Source: Chapter 2, Sections 2.3 and 2.4.2

---

• Universal indicator: red = acidic (low pH, high H⁺); blue = basic (high pH, low OH⁻).
• Green on universal indicator = pH 7 = neutral. CBSE expects you to link green colour → neutral salt → strong acid + strong base. Both justifications are essential for full marks.

(i) X is Gypsum (CaSO₄·2H₂O) and Y is Plaster of Paris (CaSO₄·½H₂O).

Reaction: CaSO₄·2H₂O $\xrightarrow{373\text{ K}}$ CaSO₄·½H₂O + 1½H₂O

(ii) X (Gypsum) can be reobtained from Y (Plaster of Paris) by adding water to it. Plaster of Paris absorbs water and sets into a hard solid mass, converting back to gypsum.

CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O

• Examiners expect you to correctly name both X and Y with their chemical formulae.
• The key fact is that heating gypsum at 373 K removes ¾ of the water of crystallisation to form Plaster of Paris (not complete dehydration).
• For part (ii), the reverse reaction — Plaster of Paris absorbing water to reform gypsum — is the key concept. Writing the chemical equation earns full marks.
• Since no source passage directly covers this topic, this answer is based on standard CBSE Class 10 Science Chapter 2 (Acids, Bases and Salts) content.

(c) Hydrogen chloride gas and water

When HCl gas dissolves in water, it ionises to form H⁺ ions, which combine with water molecules to form hydronium ions: HCl + H₂O → H₃O⁺ + Cl⁻.

Hydronium ion (H₃O⁺) is formed when H⁺ (from an acid) combines with a water molecule. HCl is an acid that dissociates in water to release H⁺, which then bonds with H₂O to give H₃O⁺. NaOH gives OH⁻ (not H₃O⁺), CaCl₂ is a salt with no such reaction, and ethanol does not ionise in water.

(d) Baking soda (Sodium hydrogen carbonate)

Fresh milk has pH 6 (slightly acidic). Adding baking soda raises the pH to slightly alkaline, which delays the formation of lactic acid (curdling). As it is a mild base, it neutralises the developing acidity.

Source: Acids, Bases and Salts, Section 2.3.1 / Exercise Q.12

---

The question is directly based on Exercise Q.12 of Chapter 2. The key logic is:

• Curdling occurs when bacteria produce lactic acid, lowering pH further.
• Adding a mild base (baking soda / NaHCO₃) shifts pH to slightly alkaline, slowing acid build-up and delaying curdling.
• Sodium carbonate (Na₂CO₃) and NaOH are too strongly basic and would spoil the milk; baking powder is a mixture, not a pure base. Only baking soda is the appropriate mild base used for this purpose.

(a) Clove oil and vanilla essence

Olfactory indicators change their smell in acidic or basic media. Clove oil and vanilla essence both show a change in odour with acids and bases, making them olfactory indicators.

The textbook (Activity 2.2, Chapter 2) states that onion, vanilla essence, and clove oil are tested as olfactory indicators. However, onion is also confirmed as one, but turmeric is a colour-based indicator. Litmus and methyl orange are synthetic colour indicators — not olfactory. The pair that consists of only olfactory indicators is clove oil and vanilla essence (option a). Students often confuse turmeric and onion — remember turmeric is a colour indicator, not olfactory.

Dry HCl gas does not ionise in the absence of water and hence does not produce H⁺ ions. Since no H⁺ (hydronium) ions are formed, it shows no acidic character and cannot change the colour of dry blue litmus paper.

$$\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$$

H⁺ ions are produced only in the presence of water. Without water, HCl remains as molecules and cannot act as an acid.

Source: Chapter 2, Section 2.2.1

---

• The key concept is that acids show acidic properties only in aqueous solution — they must ionise to release H⁺ ions.
• Dry litmus paper has no moisture, so HCl gas cannot ionise → no H₃O⁺ → no colour change.
• Always mention: (1) no ionisation without water, (2) no H⁺ ions produced, (3) therefore no acidic behaviour. Writing the equation earns a mark in most marking schemes.

When sulphuric acid (H₂SO₄) reacts with sodium hydroxide (NaOH), the salt formed is sodium sulphate (Na₂SO₄).

$$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

Since sulphuric acid is a strong acid and sodium hydroxide is a strong base, sodium sulphate is a neutral salt with pH = 7.

Source: Chapter 2, Section 2.4.2

---

• Examiners expect: name of salt, chemical equation (balanced), and the pH value with reasoning.
• Key rule to remember: Strong acid + Strong base → Neutral salt, pH = 7.
• Always justify the pH by stating the nature of the acid and base used — this earns the reasoning mark.
• Writing the balanced equation is important; note that 2 moles of NaOH are needed for H₂SO₄.

To safely dilute a strong concentrated acid, the acid must always be added slowly to water with constant stirring, and never the other way round.

Reason: Dissolving an acid in water is a highly exothermic process. If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns. The glass container may also break due to excessive local heating.

Source: Chapter 2, Section 2.2.1

---

Examiners expect two things: (1) the correct procedure — acid added to water, not water to acid — and (2) the reason involving the exothermic nature of the process and the danger of splashing/burns. Both are needed for full marks. This is a standard 2-mark question from the in-text questions of Section 2.2.1.

• X = Sodium chloride (NaCl) / brine (aqueous solution)
• Y = Hydrogen (H₂) — liberated at cathode
• Z = Chlorine (Cl₂) — liberated at anode
• B = Bleaching powder — formed when Cl₂ reacts with dry slaked lime Ca(OH)₂

$$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{electrolysis}} 2\text{NaOH} + \text{H}_2\uparrow + \text{Cl}_2\uparrow$$

This question tests the chlor-alkali process. Remember the key facts: electrolysis of brine (NaCl solution) gives NaOH (caustic soda) as the main product, H₂ at cathode, and Cl₂ at anode. Cl₂ + dry slaked lime [Ca(OH)₂] → bleaching powder. Examiners expect all four names correctly identified. Writing the equation adds value even in a 2-mark answer.

(c) Sodium carbonate decahydrate (Na₂CO₃.10H₂O)

The textbook (Section 2.4.3) explicitly states that one of the uses of washing soda (Na₂CO₃.10H₂O) is "for removing permanent hardness of water." It reacts with soluble calcium and magnesium salts causing hardness, converting them into insoluble carbonates. The other options are incorrect: NaHCO₃ is baking soda (antacid/fire extinguisher), NaCl is common salt, and CaSO₄.½H₂O is Plaster of Paris.

Source: Chapter 2, Section 2.4.3

(b) Sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.

The textbook (Chapter 2, Section 2.2.1) explicitly states: "Bases which are soluble in water are called alkalis." So the defining criterion for an alkali is water-solubility, not strength. NaOH dissolves in water → alkali; Fe(OH)₃ does not dissolve in water → base but not an alkali. Option (a) is partially correct but solubility, not strength, is the textbook definition. Option (d) is wrong in reasoning — "comparatively higher solubility" misses the point.

(d) Oxalic acid

Table 2.3 ("Some naturally occurring acids") in Chapter 2 directly lists Tomato → Oxalic acid. Students must memorise this table as MCQs frequently test it. Note: Acetic acid is from vinegar, Lactic acid from curd, and Methanoic acid from ant/nettle sting.

(c) Carbon dioxide; it turns lime water milky.

NaHCO₃ + HCl → NaCl + H₂O + CO₂. CO₂ is evolved, which turns lime water milky.

When an acid reacts with a metal hydrogencarbonate (bicarbonate), it produces CO₂ gas (not hydrogen). CO₂ turns lime water milky — this is its standard test. Option (d) is wrong because CO₂ does not cause a pop sound; that test is for hydrogen gas.

(a)

• (i) Less than 7 (acidic): Group A and Group B — lemon juice, vinegar, aerated drink, tomato juice, coffee, and ginger juice are all acidic.
• (ii) Greater than 7 (basic): Group C (partially) — sodium hydroxide (pH ≈ 14) and lime water (pH > 7) are basic. Sodium chloride is neutral (pH = 7).

(b) Two ways to determine pH:

1. Using pH paper (universal indicator paper) — colour change indicates pH.
2. Using universal indicator solution — different colours correspond to different pH values.

(c) Copper vessels tarnish due to the formation of copper oxide (CuO) on their surface. Sour substances like lemon juice contain acids (citric acid) which react with copper oxide, dissolving it and restoring the shiny surface of the copper vessel.

Source: Chapter 2 – Acids, Bases and Salts, NCERT Science Class 10

---

• (a): Acids have pH < 7, bases have pH > 7, neutral = 7. All items in Groups A and B are acidic; NaOH and lime water are basic; NaCl is neutral.
• (b): Examiners expect "pH paper/universal indicator paper" and "universal indicator solution" as the two standard methods. Digital pH meter is also acceptable.
• (c): Key idea: tarnish = metal oxide layer; acid reacts with metal oxide to remove it. Mention the specific acid (citric acid) and the product (copper oxide dissolves). This is the standard explanation expected in board exams.

(d) (ii), (iii) and (iv)

Wait — correct answer is (c) (i) and (ii).

The bulb glows only with dilute HCl (i) and dilute H₂SO₄ (ii), as acids produce ions in solution that conduct electricity. Glucose and alcohol do not ionise, so the bulb does not glow with them.

Source: Chapter 2, Activity 2.8

• Examiners expect you to recall that only acids and bases produce ions in aqueous solution, enabling electrical conduction.
• Glucose and alcohol contain hydrogen but do not ionise in water, so they cannot conduct electricity.
• The key concept tested: electrical conductivity requires free ions — acids provide H⁺ and anions; non-electrolytes like glucose/alcohol do not.
• Option (c) is correct because both HCl and H₂SO₄ are strong acids that fully dissociate into ions.

The clues identify:

• X = Tartaric acid (acid in tamarind) — Chemical formula: C₄H₆O₆
• Y = Baking soda (base mixed into flour) — Common name: Baking soda; Chemical formula: NaHCO₃
• Z = Washing soda (prepared from common salt, used for faster cooking) — Common name: Washing soda; Chemical formula: Na₂CO₃

(i) X = Tartaric acid, C₄H₆O₆; Y = Baking soda, NaHCO₃; Z = Washing soda, Na₂CO₃

(ii) Baking soda (NaHCO₃) is prepared by treating a solution of sodium chloride (brine) with ammonia and CO₂ (Solvay process), or from washing soda and CO₂:

$$\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow 2\text{NaHCO}_3$$

On heating, NaHCO₃ releases CO₂ gas:

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

The CO₂ produced gets trapped in the dough, making the cake soft and spongy.

(iii) Milk of magnesia — Mg(OH)₂ — is a mild base used as an antacid.

Source: Chapter 2, Salts and their uses

---

• The question is a disguised identification puzzle: tamarind acid = tartaric acid (X), the base added to flour = baking soda (Y), prepared from common salt and used for faster cooking = washing soda (Z).
• Examiners expect all three names + formulae in (i); don't skip any.
• For (ii), the key points are: how NaHCO₃ is made and the decomposition equation showing CO₂ release — both are required for full marks.
• For (iii), Mg(OH)₂ (milk of magnesia) is the standard NCERT answer for a mild base antacid. Do not write NaOH (it is a strong base).

(i) FeSO₄ crystals exist as hydrated salt — FeSO₄·7H₂O. On heating, the water of crystallisation is driven off, which condenses as tiny droplets on the cooler upper walls of the boiling tube.

(ii) The green colour of ferrous sulphate crystals changes to reddish-brown, as FeSO₄ decomposes to form ferric oxide (Fe₂O₃).

$$2\text{FeSO}_4 \xrightarrow{\Delta} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$$

(iii) 7 molecules of water are attached per molecule of FeSO₄.

• (I) Copper sulphate: CuSO₄·5H₂O
• (II) Sodium carbonate: Na₂CO₃·10H₂O

(iv) Plaster of Paris is obtained by heating gypsum (CaSO₄·2H₂O) at 373 K:

$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$

Uses: (1) Making casts for fractured bones. (2) Making statues and decorative materials.

Source: Chapter 1, Section 1.2.2; Chapter 2 (Acids, Bases and Salts)

---

• (i) Key phrase: "water of crystallisation" — examiners expect this term.
• (ii) The colour change (green → reddish-brown) is due to formation of Fe₂O₃; mention both colours.
• (iii) The 7 water molecules must be stated clearly. Both crystal formulae carry 1 mark — memorise them accurately (CuSO₄·5H₂O and Na₂CO₃·10H₂O).
• (iv) This sub-question is from the Acids, Bases and Salts chapter (Ch. 2). State the temperature (373 K), write the equation, and give two distinct uses for full marks.

(C) 7

Sodium chloride (NaCl) is a salt of strong acid (HCl) and strong base (NaOH). Salts of a strong acid and a strong base are neutral with pH value of 7.

Source: Chapter 2, Section 2.4.2

---

The key rule to remember: strong acid + strong base → neutral salt (pH = 7). The textbook explicitly states this in Section 2.4.2 and also describes NaCl as a neutral salt in Section 2.4.3. For MCQs like this, eliminate options below 7 (acidic) and above 7 (basic), since NaCl neither hydrolyses to give H⁺ nor OH⁻ in excess.

(A) Calcium phosphate

Tooth enamel is made up of calcium phosphate, which is the hardest substance in the human body.

This is a standard factual recall question from the Science textbook (Class 10). The source passage mentions that bacteria produce acids which demineralise the enamel (calcium phosphate). Calcium phosphate is the salt that constitutes tooth enamel — remember it as a calcium salt of phosphoric acid. The other options (Mg, Na, Al phosphate) are distractors; none of these form tooth enamel.

Common name: Plaster of Paris
Chemical name: Calcium sulphate hemihydrate

Preparation: Plaster of Paris is prepared by heating gypsum (CaSO₄·2H₂O) at 373 K (100°C).

$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O}$$

Reaction with water:

When water is added to Plaster of Paris, it sets into a hard solid mass — gypsum.

$$\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O} \rightarrow \text{CaSO}_4 \cdot 2\text{H}_2\text{O}$$

---

• Examiners expect all three parts: name, preparation with equation, and the reverse reaction equation.
• Note that Plaster of Paris sets by recombining with water to reform gypsum — this is the reverse of its preparation, so knowing both equations together is essential.
• The formula CaSO₄·½H₂O (or CaSO₄·¼H₂O in some books) — NCERT Class 10 Science uses the ½H₂O form; always match your textbook edition.
• The temperature 373 K is a key detail examiners look for.

(a)
(i) The gas evolved is hydrogen chloride (HCl) gas.

(ii)

• (I) Dry blue litmus paper: No change in colour observed.
• (II) Wet blue litmus paper: Turns red.

Conclusion: HCl gas is acidic in nature. It shows acidic character only in the presence of water (moisture), as it dissolves in water to form hydrochloric acid (H⁺ ions). In dry state, it does not ionise and hence shows no acidic behaviour.

---

The key concept here is that HCl is a covalent compound and shows acidic nature only when dissolved in water — it ionises to give H⁺ ions only in the presence of moisture. Dry litmus test vs. wet litmus test distinguishes this beautifully. Examiners expect you to state both observations clearly and draw a conclusion about the role of water in acidic behaviour. This is a standard practical-based question from the Acids, Bases and Salts chapter (Chapter 2).

Option D: Hydrochloric acid and sulphuric acid.

Acids produce H⁺(aq) ions in solution, which carry electric current and make the bulb glow. Glucose and alcohol do not ionise, so they do not conduct electricity.

Source: Chapter 2, Section 2.2 (Activity 2.8)

The key concept is ionic conduction: only solutions containing ions can conduct electricity. HCl and H₂SO₄ are strong acids that fully dissociate into ions in water. Glucose and alcohol contain hydrogen but do not produce H⁺ ions in solution, so they are non-electrolytes. Examiners expect you to link "glowing bulb = ions present = acid/base" directly.

Answer: D

HCl (strong acid) + NH₄OH (weak base) → NH₄Cl (acidic salt, pH < 7), which turns blue litmus red.

Source: Acids, Bases and Salts, Section 2.4.2

• A salt turns blue litmus red only if it is acidic (pH < 7).
• Salts of strong acid + weak base are acidic.
• Option D: HCl (strong acid) + NH₄OH (weak base) → NH₄Cl — acidic salt ✓
• Options A & B: strong acid + strong base → neutral salt (no litmus change).
• Option C: CH₃COOH (weak acid) + NaOH (strong base) → basic salt (turns red litmus blue, not the other way).

Experimental Set-up (Activity 2.8):

• Take solutions of glucose, alcohol, HCl, and H₂SO₄ separately.
• Fix two nails on a cork placed in a beaker. Connect nails to a 6V battery through a bulb and switch.
• Pour each solution into the beaker and switch on the current.

Observation: The bulb glows with HCl and H₂SO₄ solutions but does not glow with glucose and alcohol solutions.

Reason: Acids produce H⁺(aq) ions in water, which conduct electricity. Glucose and alcohol do not ionise in water — they do not produce H⁺ ions — and hence are not classified as acids, even though they contain hydrogen.

Source: Chapter 2, Section 2.2

---

• Examiners expect: (1) the experimental setup description, (2) observation (bulb glows/doesn't glow), and (3) the reason linking ion formation to acidic character.
• Key point: containing hydrogen ≠ being an acid. Only compounds that release H⁺(aq) ions in solution are acids.
• Avoid writing unnecessary details about glucose/alcohol chemical formulas — focus on ion formation.

(i) Sodium Hydroxide:
An aqueous solution of sodium chloride (brine) is electrolysed. This is called the chlor-alkali process. NaOH is formed near the cathode, Cl₂ at the anode, and H₂ at the cathode.

$$2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{Cl}_2\text{(g)} + \text{H}_2\text{(g)}$$

(ii) Sodium Hydrogen Carbonate:
NaCl solution is treated with water, CO₂, and NH₃. The less soluble NaHCO₃ is precipitated.

$$\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$$

Source: Acids, Bases and Salts, Section 2.4.3

---

• Examiners expect the name of the process (chlor-alkali) for NaOH — this is a common 1-mark component.
• Both balanced equations must be written correctly; missing state symbols may cost half a mark.
• For NaHCO₃, mention all four reactants (NaCl, H₂O, CO₂, NH₃) — students often forget NH₃.
• Keep descriptions brief; the equation carries most of the marks.

(a)
(i) Acids generate H⁺ (hydrogen ions / hydronium ions, H₃O⁺) in water.
(ii) Bases generate OH⁻ (hydroxide ions) in water.

(b)
Dry HCl gas does not ionise in the absence of water and thus produces no H⁺ ions. Since litmus changes colour only in the presence of H⁺ ions (in solution), dry HCl cannot affect dry litmus paper.

---

• For (a), examiners expect the correct ionic formulae: H⁺/H₃O⁺ for acids and OH⁻ for bases. Either H⁺ or H₃O⁺ is accepted.
• For (b), the key concept is that ionisation requires water as a medium. Without water, HCl remains molecular and cannot produce H⁺ ions, so no colour change occurs. Always link the absence of water → no ions → no effect on litmus.

(B) K₂SO₄, Na₂SO₄, CaSO₄

These three salts all contain the same negative radical (SO₄²⁻), so they belong to the family of sulphate salts.

The textbook defines a family of salts as salts sharing the same positive or negative radical. Option B has K⁺, Na⁺, and Ca²⁺ as different positive radicals but all share SO₄²⁻ — making them the sulphate family. Option A (NaCl, Na₂SO₄, CaSO₄) mixes chloride and sulphate anions, so it is not a single family. Always check that ALL salts in the option share one common radical.

(a) Solution P (strong acid) would be placed near pH 0–3 (towards the left/acidic end), and Solution Q (strong base) would be placed near pH 11–14 (towards the right/basic end) of the pH scale.

(b)
(i) To increase the pH (make it more basic): NaOH (sodium hydroxide)
(ii) To decrease the pH (make it more acidic): HCl (hydrochloric acid)

(c)(i) When pH decreases from 4 to 2, the hydronium ion (H₃O⁺) concentration increases 100 times (increases by a factor of 100).
Colour change on pH paper: from orange (pH 4) to red (pH 2).

Source: Chapter 2 – Acids, Bases and Salts; pH Scale section

---

• (a) Strong acids have very low pH (0–3); strong bases have very high pH (11–14). Place them at opposite ends of the scale.
• (b) Any valid strong base/acid is acceptable. NaOH and HCl are the most common NCERT examples.
• (c) pH is a logarithmic scale — a drop of 2 units means H⁺ concentration increases 100-fold (10² = 100). Examiners expect "100 times" specifically. Colour change from orange → red is read directly from the pH scale figure: pH 4 ≈ orange, pH 2 ≈ red.

(D) 7.0 to 7.8

Our body works within the pH range of 7.0 to 7.8.

Source: Acids, Bases and Salts, Section 2.3.1

---

The passage explicitly states: "Our body works within the pH range of 7.0 to 7.8." This is a direct fact to memorise. Note that this range is slightly alkaline (above 7), which reflects the slightly basic nature of blood and body fluids. Options A and B are too acidic; option D is too alkaline.

(C) Sodium

The corrosive warning sign is displayed on containers of sodium hydroxide (NaOH) pellets, as it is a strongly corrosive alkali.

Source: Chapter 2, Section 2.2.1 (Fig. 2.5)

The textbook explicitly mentions the corrosive warning sign (Fig. 2.5) in the context of concentrated sulphuric acid and sodium hydroxide pellets. Among the four options, only sodium hydroxide (NaOH) is referenced with this warning sign. Ca(OH)₂, Mg(OH)₂, and Al(OH)₃ are either insoluble, weak, or amphoteric — not strongly corrosive alkalis that require this label.

(a) Electrolysis of brine (aqueous NaCl solution):

$$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$$

(b) Uses of Chlorine (Cl₂):

1. Used as a disinfectant for purification of drinking water.
2. Used in the manufacture of PVC and disinfectants.

(Or uses of NaOH: used in making soap; used in paper industry.)

(c) Brine is a concentrated aqueous solution of common salt (NaCl). When electricity is passed through brine, it undergoes electrolysis to produce three important chemicals — sodium hydroxide (NaOH), chlorine gas, and hydrogen gas. This process is called the chlor-alkali process. These products are widely used in industries such as soap-making, textile, and water treatment.

---

• (a) Examiners expect a fully balanced equation with state symbols — don't skip those.
• (b) Name ONE product clearly and give TWO distinct uses. Chlorine or NaOH are the safest choices as their uses are explicitly in the textbook.
• (c) Sub-question (c) appears to be cut off in the original, so the answer covers the chlor-alkali process — the most likely intended topic given the passage. In the exam, answer whatever the actual question asks. Key terms: chlor-alkali process, brine, electrolysis — examiners look for these.

(A) Salt and water is formed.

• HCl + NaOH → NaCl + H₂O (neutralisation)
• MgO + 2HCl → MgCl₂ + H₂O (metal oxide + acid)
• CO₂ + 2NaOH → Na₂CO₃ + H₂O (non-metallic oxide + base)

In each case, a salt and water are produced.

The key concept tested is that acid-base reactions, metal oxide + acid reactions, and non-metallic oxide + base reactions all yield salt + water. Option B is wrong because Na₂CO₃ is not a neutral salt. Option C is wrong — no hydrogen gas is released in any of these three reactions. Option D is incorrect as acidic salts are not necessarily formed here.

(D) Weak acid, neutral, strong base, strong acid

From the pH scale: A is at pH 12 (strong base), B is at pH 5 (weak acid), C is at pH 6 (neutral/near-neutral but slightly acidic — effectively neutral here), and D is at pH 3 (strong acid). The order asked is A, B, C, D → Strong base, weak acid, neutral, strong acid — which matches option (D).

Source: Chapter 2, Section 2.3

Read the order carefully — the question asks for A, B, C, D respectively. From the diagram: A ≈ pH 12 (strongly basic), B ≈ pH 5 (weakly acidic, below 7), C ≈ pH 6 (close to neutral/treated as neutral in this context), D ≈ pH 3 (strongly acidic). Match these to the options in the same A→B→C→D sequence to get option (D).

(i) NaOH by Chlor-Alkali Process:

When electricity is passed through aqueous NaCl solution (brine), it decomposes to give NaOH, chlorine gas, and hydrogen gas:

$$2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \xrightarrow{\text{electricity}} 2\text{NaOH(aq)} + \text{Cl}_2\text{(g)} + \text{H}_2\text{(g)}$$

(ii) Baking Soda (NaHCO₃):

Baking soda is prepared by passing CO₂ through a concentrated solution of sodium chloride saturated with ammonia (Solvay process):

$$\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$$

(iii) Plaster of Paris (CaSO₄·½H₂O):

It is prepared by heating gypsum (CaSO₄·2H₂O) at 373 K:

$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$

---

• These three equations are directly from the CBSE Class 10 Chapter on Acids, Bases and Salts.
• Chlor-alkali process is so named because products are chlorine (chlor) and alkali (NaOH). The electrolysis equation must be balanced and include water as a reactant.
• Baking soda: The Solvay process equation is standard; NaHCO₃ precipitates out as it is less soluble.
• Plaster of Paris: Remember it is heated at exactly 373 K (not higher), and the formula CaSO₄·½H₂O (hemihydrate) must be written correctly — examiners check this.

(i) When concentrated acid is mixed with water, a large amount of heat is released. If water is added to acid, the heat generated can cause the mixture to splash and cause burns. Adding acid to water ensures the heat is absorbed by the large volume of water gradually, making it safe.

(ii) Acids produce H⁺ (or H₃O⁺) ions only in the presence of water. Dry HCl gas has no water, so it cannot produce H⁺ ions and does not show acidic properties. Dilute HCl is dissolved in water, producing H⁺ ions, which turn blue litmus red.

(iii) Copper sulphate crystals (CuSO₄·5H₂O) contain water of crystallisation. On heating, this water is lost, converting the blue hydrated salt into white anhydrous copper sulphate (CuSO₄).

---

• These are standard reasons from the Acids, Bases and Salts chapter (Chapter 2, Science Class 10).
• For (i), the key word is safety and heat absorption by water.
• For (ii), the key concept is ionisation requires water — no water = no H⁺ ions = no acidic effect.
• For (iii), the key term is water of crystallisation — its removal causes the colour change. Always name both forms of the compound.
• Note: The source passages provided were unrelated to this question; the answers are grounded in standard NCERT Class 10 Chemistry content.

(a) Gas X = Chlorine (Cl₂); Gas Y = Hydrogen (H₂)

(b) Electrolysis of brine (aqueous NaCl) produces sodium hydroxide (Z):

$$2NaCl + 2H_2O \xrightarrow{\text{electricity}} 2NaOH + Cl_2 + H_2$$

(c) (i) or (ii) — answered as (i): Name substance 'A' and write its chemical equation.

When chlorine gas (X) is passed through slaked lime [Ca(OH)₂], bleaching powder is formed as substance 'A'.

$$2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O$$

(ii) Uses of substance 'Z' (NaOH): It is used in making soap, paper, and as a drain cleaner.

Source: Chapter 2 — Acids, Bases and Salts; Electrolysis of Brine / Chemicals from Common Salt

---

• X = Chlorine because it disinfects water and reacts with slaked lime. Y = Hydrogen because it burns with a pop sound — classic test for H₂.
• The balanced equation for (b) is the standard electrolysis of brine equation — examiners award the mark only if it is balanced with correct formulae.
• For (c), bleaching powder is Ca(OCl)Cl (calcium oxychloride); examiners accept either formula. If the sub-question asked for uses of NaOH instead, soap/paper/drain cleaner are the three accepted uses from the textbook.
• Always write chemical names alongside formulae to avoid losing marks for ambiguity.

(C) Sodium hydrogen carbonate + Tartaric acid

Baking powder is a mixture of sodium hydrogen carbonate (NaHCO₃) and a mild edible acid like tartaric acid. When heated or mixed with water, the acid reacts with NaHCO₃ to release CO₂, which makes the baked goods rise. Options with sodium carbonate or acetic acid are incorrect.

(A) Less than 7

Gastric glands release hydrochloric acid (HCl) during digestion, making the gastric juice acidic. Any acidic solution has a pH less than 7.

The passage states that gastric glands release hydrochloric acid, which creates an acidic medium for pepsin to act. Acidic solutions always have pH < 7. Option (D) — pH = 0 — is incorrect as pure HCl is pH 0 but gastric juice is dilute acid, not pure acid. Examiners expect you to simply select (A) and may ask for a one-line justification.

(a) The acid present in ant's sting is methanoic acid (formic acid), HCOOH.

(b)
(i) Dissolving acid in water is highly exothermic. If water is added to acid, the heat generated is so intense that it may cause the mixture to splash and cause burns. Adding acid to water ensures the heat is absorbed by the large volume of water, making the process safer.

(ii) Baking soda (NaHCO₃) is a mild base. When excess acid is produced in the stomach causing acidity, baking soda neutralises it:
NaHCO₃ + HCl → NaCl + H₂O + CO₂
Thus it acts as an effective antacid.

---

• (a) is a direct 1-mark recall fact — just naming the acid is enough.
• (b)(i) Examiners want the key phrase "exothermic reaction" and the safety reason (splashing/burns). Mentioning that water absorbs the heat is key.
• (b)(ii) Must state that NaHCO₃ is a mild base and show/mention the neutralisation reaction. Avoid saying it is a strong base — that would be wrong.

(a) Plaster of Paris is calcium sulphate hemihydrate.
Formula: $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$

(b) It is prepared by heating gypsum ($\text{CaSO}_4 \cdot 2\text{H}_2\text{O}$) at 373 K:

$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$

(c) Two uses of Plaster of Paris:

1. Used by doctors for setting fractured bones.
2. Used for making toys, decorative materials, and smooth surfaces in construction.

---

• (a) Always write both the name and formula; the formula with ½ water of crystallisation is essential.
• (b) The key detail is the temperature 373 K (not boiling point). Writing gypsum's formula correctly on the left earns marks.
• (c) Any two valid uses are accepted. Medical (bone-setting) and decorative/construction are the standard NCERT examples to memorise.

(i) Olfactory indicator: Substances whose odour changes in acidic or basic media are called olfactory indicators. Examples: onion, vanilla essence, clove oil.

(ii) Water of crystallisation: The fixed number of water molecules present in one formula unit of a salt is called water of crystallisation. Example: CuSO₄·5H₂O.

Source: Acids, Bases and Salts, sections 2.1.1 and "What you have learnt"

For olfactory indicators, the key phrase from the textbook is "substances whose odour changes in acidic or basic media." Always give an example.

For water of crystallisation, copy the textbook definition exactly — "fixed number of water molecules present in one formula unit of a salt." An example (like blue vitriol) earns full marks. Since this is a 1-mark question split into two parts, each definition needs just one concise line.

(i) Olfactory indicator: A substance whose odour changes in acidic or basic media, e.g., onion, vanilla essence, clove oil.

(ii) Water of crystallisation: The fixed number of water molecules chemically bonded in the crystalline structure of a salt, e.g., CuSO₄·5H₂O.

Source: Chapter 2, Section 2.1.1

The question asks for two definitions but is worth 1 mark total — keep each definition to one short phrase. Olfactory indicator is directly defined in the passage. Water of crystallisation is a standard textbook definition from Chapter 2 (not in the given passages, but is a core term from the same chapter). Examiners expect the definition plus one example for each.

(C) Hydrogen

Magnesium (Mg) reacts with very dilute HNO₃ to evolve H₂ gas, unlike other metals where HNO₃ (a strong oxidising agent) oxidises H₂ to water.

The textbook explicitly states: "But magnesium (Mg) and manganese (Mn) react with very dilute HNO₃ to evolve H₂ gas." This is an exception to the general rule that metals do not produce hydrogen with nitric acid. Examiners expect students to recall this specific exception.

(D) hydrochloric acid and sodium hydroxide

HCl and NaOH ionize in water to produce H⁺ and OH⁻ ions respectively, allowing current to flow and the bulb to glow. Fructose and ethanol do not ionize, so they do not conduct electricity.

Source: Chapter 2, Section 2.2

The key concept is ionic conduction: only solutions that produce free ions conduct electricity. Acids produce H⁺(aq) and bases produce OH⁻(aq) — both allow current flow. Non-electrolytes like fructose (sugar) and ethanol have no ions in solution, so the bulb stays off. Examiners expect you to identify the two ionic solutions correctly.

(i) Baking Soda (NaHCO₃):
Baking soda is prepared by passing excess CO₂ through a saturated solution of sodium chloride (brine) in the presence of ammonia:

$$\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \rightarrow \text{NaHCO}_3 + \text{NH}_4\text{Cl}$$

(ii) Bleaching Powder (CaOCl₂):
Bleaching powder is prepared by passing chlorine gas over dry slaked lime [Ca(OH)₂]:

$$\text{Ca(OH)}_2 + \text{Cl}_2 \rightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$$

(iii) Plaster of Paris (CaSO₄·½H₂O):
Plaster of Paris is prepared by heating gypsum (CaSO₄·2H₂O) at 373 K:

$$\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O}$$

Source: Acids, Bases and Salts, Chapter 2

---

• All three are standard CBSE questions from Chapter 2 (Acids, Bases and Salts). Each preparation must include the correct equation — examiners award marks for the balanced equation specifically.
• For baking soda, name the process (Solvay-type) and mention CO₂ + NH₃ + brine.
• For bleaching powder, remember it is Cl₂ over dry slaked lime.
• For Plaster of Paris, the temperature 373 K (100°C) is important — higher temperature gives anhydrite (dead burnt plaster), so mention it.
• Writing ½H₂O in the formula of Plaster of Paris is acceptable; some write it as CaSO₄·½H₂O.

Chlor-Alkali Process:

When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to give sodium hydroxide, chlorine, and hydrogen. This is called the chlor-alkali process.

$$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{electricity}} 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$$

• At Anode: Chlorine gas (Cl₂) is produced.
• At Cathode: Hydrogen gas (H₂) is produced; sodium hydroxide (NaOH) is formed near the cathode.

Source: Acids, Bases and Salts, Chapter 2

---

• The name "chlor-alkali" comes from the two main products: chlorine (chlor) and alkali (NaOH).
• Examiners expect: (1) a clear definition/description of the process, (2) the balanced chemical equation with state symbols, and (3) specific products named at each electrode.
• Remember: Cl₂ → anode; H₂ → cathode; NaOH forms near cathode. All three products must be mentioned for full marks.

ZnO — amphoteric oxide; Na₂O — basic oxide; CO₂ — acidic oxide.

Examiners expect all three categorised correctly in one line. ZnO reacts with both acids and bases (amphoteric); Na₂O is a metal oxide (basic); CO₂ is a non-metal oxide (acidic). No explanation needed for 1 mark.

Option A: tomato, curd, ant-sting

Oxalic acid is found in tomato, lactic acid in curd, and methanoic acid (formic acid) in ant-sting.

This is standard factual knowledge from the NCERT science curriculum. Methanoic acid = formic acid (found in ant/bee stings and nettle sting — but the specific pairing here: oxalic acid → tomato, lactic acid → curd, methanoic acid → ant-sting matches Option A). Note: nettle sting contains methanoic acid too, but Option A is the standard accepted answer matching all three correctly as per NCERT context.

Answer: B — Vanilla essence

Vanilla essence is an olfactory indicator — its odour changes in acidic/basic media, making it suitable for visually impaired students who cannot detect colour changes.

• Visually impaired students cannot observe colour changes, so visual indicators (turmeric, methyl orange, litmus) are not useful for them.
• Olfactory indicators work by change in smell, not colour. The textbook (Activity 2.2) identifies vanilla essence as one such olfactory indicator — its characteristic odour is lost in a basic (NaOH) solution but retained in an acidic one.
• Onion and clove oil are also olfactory indicators mentioned in the chapter, but vanilla essence is the one given in the options.

(i)

I. CH₃–CH₂–C(=O)–H

The compound has 3 carbon atoms with an aldehyde (–CHO) functional group.
Name: Propanal

II. CH₃–C≡CH

The compound has 3 carbon atoms with a triple bond (alkyne).
Name: Propyne

(ii)

I. The sweet-smelling compounds produced belong to the class Esters.

II. The reaction is called Esterification. The chemical equation is:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(Ethanoic acid + Ethanol → Ethyl ethanoate + Water)

(iii)

Soap does not lather with hard water because hard water contains dissolved salts of calcium and magnesium. The soap reacts with these salts to form an insoluble precipitate called scum (calcium or magnesium salt of the fatty acid), which reduces the soap available for cleaning and prevents foam formation.

Source: Carbon and its Compounds, 4.2.5 Nomenclature of Carbon Compounds

---

• (i): Count the carbon atoms → assign the parent chain name. Aldehyde suffix = -al; triple bond suffix = -yne. Common errors: writing "propanaldehyde" or "methylacetylene" — CBSE expects IUPAC names.
• (ii): Always write the full balanced equation with the acid catalyst shown above the arrow. State the class name clearly as "Esters."
• (iii): The key word examiners look for is scum and the explanation that Ca²⁺/Mg²⁺ ions react with soap to form insoluble salts, consuming the soap. One mark — keep it to 2–3 lines.

(i) Carbon cannot form C⁴⁺ or C⁴⁻:

To form C⁴⁺, carbon would need to remove 4 electrons, requiring a very large amount of energy, which is not energetically feasible. To form C⁴⁻, it would need to gain 4 electrons; placing 4 extra electrons on a small carbon nucleus would be very difficult as the nucleus cannot hold so many electrons with stability. Therefore, carbon shares electrons and forms covalent bonds instead.

(ii) Homologous Series:

A homologous series is a group of organic compounds having the same general formula and similar chemical properties, differing from each other by a –CH₂– unit.

C₄H₁₀ has more carbon atoms and hence greater molecular mass than C₃H₈ or C₂H₆. Greater molecular mass means stronger intermolecular forces (van der Waals forces), so more energy is needed to overcome them, resulting in higher melting and boiling points.

(iii) Oxidation of Ethanol:

When ethanol is heated with acidified potassium dichromate, it gets oxidised to ethanoic acid (acetic acid). The orange colour of potassium dichromate changes to green.

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$$

---

• (i) Examiners want both reasons — energy cost of losing 4e⁻ AND instability of gaining 4e⁻. Missing either loses a mark.
• (ii) Definition must include: same general formula, differ by –CH₂–, similar chemical properties. For boiling/melting point, link molecular mass → intermolecular forces → energy required.
• (iii) 1-mark answer needs: product (ethanoic acid) + colour change of dichromate (orange → green), or the equation. Keep it to one or two lines.

(a) Universal indicator is used to predict how strong an acid or base is. It shows different colours at different pH values, allowing us to determine the strength of an acid or base.

(b) Solution A (pH = 3) has the highest concentration of H⁺ ions, as lower pH indicates higher hydrogen ion concentration.

(c) Strong acids are acids that completely dissociate in water to produce a large number of H⁺ ions. Example: HCl, H₂SO₄.

Weak acids are acids that partially dissociate in water, producing fewer H⁺ ions. Example: acetic acid (CH₃COOH), carbonic acid.

Source: Chapter 2 – Acids, Bases and Salts

---

• (a) Universal indicator is the standard answer; examiners also accept "pH paper."
• (b) Remember: lower pH = higher H⁺ concentration. This is a direct 1-mark factual point — state the solution and the reason.
• (c) For 2 marks, you need both definitions with at least one example each. The key distinction is complete vs. partial dissociation. Do not write long paragraphs — two crisp definitions suffice.

(i) Extraction of Zn from ZnS:

Step 1 – Roasting (ZnS converted to ZnO):

$$2\text{ZnS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{ZnO} + 2\text{SO}_2$$

Step 2 – Reduction (ZnO reduced by coke):

$$2\text{ZnO} + \text{C} \xrightarrow{\Delta} 2\text{Zn} + \text{CO}_2$$

(ii) Nitric acid is a strong oxidising agent. It oxidises the hydrogen gas produced during the reaction to water (H₂O). Therefore, hydrogen gas is not evolved; instead, oxides of nitrogen (like NO, NO₂) are released.

---

• For (i), the two-step process — roasting followed by reduction — must be shown as separate balanced equations. Both steps carry marks; missing either loses a mark.
• For (ii), the key phrase is oxidising agent: HNO₃ oxidises H₂ to H₂O before it can escape, so NO/NO₂ is evolved instead. Always mention this in one clear sentence.

The chemical formula of Plaster of Paris is CaSO₄·½H₂O. Gypsum (CaSO₄·2H₂O) is heated at 373 K (100°C) to obtain Plaster of Paris.

Examiners expect both the chemical formula and the exact temperature. The formula is sometimes written as CaSO₄·½H₂O or (CaSO₄)₂·H₂O — both are accepted. The temperature 373 K is the key fact; writing only "100°C" is also acceptable. Since this is 1 mark, one concise line covering both parts is sufficient.

A = Sodium bicarbonate (NaHCO₃) — used in soda-lime fire extinguishers.
B = Sodium carbonate (Na₂CO₃) — formed on heating A; used to remove permanent hardness of water.

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$$

The key clues are: (1) used in soda-acid fire extinguisher → NaHCO₃; (2) decomposes on heating → Na₂CO₃ (washing soda); (3) washing soda removes permanent hardness by precipitating Ca²⁺/Mg²⁺ ions. Examiners expect both compounds identified with correct formula.

(A) By adding acid to water with constant stirring.

When acid is added to water slowly with constant stirring, the heat generated is absorbed by the large volume of water, preventing splashing or burns.

The source passage (Chapter 2, Section 2.2.1) explicitly states: "The acid must always be added slowly to water with constant stirring. If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns." This is a standard safety rule — remember the mnemonic: "Do as you oughta, add acid to water." Option (B) is the dangerous reverse procedure; options (C) and (D) are irrelevant to dilution safety.

(B) Hydrochloric acid

The solution is basic (turns red litmus blue). Adding excess HCl (acid) will neutralise the base, making it acidic, turning blue litmus red again — reversing the change.

• A solution that turns red litmus blue is basic/alkaline.
• To reverse this (make it acidic again), an acid must be added in excess.
• Options A, C, and D (NaOH, baking soda, slaked lime) are all basic — they would not reverse the change.
• Only HCl (option B) is an acid that can neutralise the base and make the solution acidic, turning blue litmus back to red.