Class 9 · ChemistrySet 3 · Answer key

Answer Key – Set 3 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 3, for teachers and parents to value the paper. Accept any correct answer in the student’s own words; give method marks even when the final number is wrong.

ANSWER KEY & MARKING SCHEME · SET 3

Standard IX · Chemistry · Annual Examination

Total score: 60 · Time: 2 hours

Section AAnswer all questions · 10 × 1 = 10 marks

The mass of an electron is about

(a) 1837 times the mass of a proton(b) 1/1837 of the mass of a proton(c) equal to the mass of a proton(d) equal to the mass of a neutron

[1]

Step-by-step solution

Table 1.2: mass of a proton = 1.00727 u; mass of an electron = 0.000548 u.
So the electron is about 1/1837 of the mass of a proton – so small that it is taken as 0 in practice.

✅ Answer: (b) 1/1837 of the mass of a proton

Marking: 1 mark for (b).

📖 Revise: Ch 1 · Masses of the subatomic particles · Chapter 1 – Structure of the Atom

Find the relation and fill in: iodine-131 : study and treatment of the thyroid gland :: iron-59 :

[1]

Step-by-step solution

Table 1.6 (uses of radioactive isotopes): iodine-131 – thyroid gland; cobalt-60 – cancer treatment; sodium-24 – pipeline leaks; uranium-235 – nuclear fuel.
Iron-59 is used for diagnosing anaemia.

✅ Answer: diagnosing anaemia

Marking: 1 mark.

📖 Revise: Ch 1 · Uses of isotopes – Table 1.6 · Chapter 1 – Structure of the Atom

Lanthanum (Z = 57) and the 14 elements that come after it are placed in a separate row below the periodic table. They are called

(a) actinoids(b) transition elements(c) lanthanoids(d) noble gases

[1]

Step-by-step solution

La (57) to Lu (71) are the lanthanoids (ലാൻഥനോയ്ഡുകൾ), also called rare earths. They belong to period 6.
Actinium (89) and the next 14 elements are the actinoids. Together they are the inner transition elements.

✅ Answer: (c) lanthanoids

Marking: 1 mark for (c).

📖 Revise: Ch 2 · Lanthanoids and actinoids · Chapter 2 – Periodic Table

The basicity of phosphoric acid, H₃PO₄, is

(a) 1(b) 2(c) 3(d) 4

[1]

Step-by-step solution

Basicity (ബേസികത) = the number of H⁺ ions one molecule of the acid gives.
H₃PO₄ gives 3H⁺ and PO₄³⁻, so its basicity is 3 (HCl is monobasic, H₂SO₄ dibasic).

✅ Answer: (c) 3

Marking: 1 mark for (c).

📖 Revise: Ch 3 · Acids and basicity – Table 3.8 · Chapter 3 – Chemical Bonding

Pick the odd one out and give the reason: NaOH, KOH, HNO₃, Ca(OH)₂

[1]

Step-by-step solution

NaOH, KOH and Ca(OH)₂ are bases (alkalis): they give OH⁻ ions.
HNO₃ (nitric acid) gives H⁺ ions – it is an acid. So HNO₃ is the odd one out.

✅ Answer: HNO₃ – it is an acid (gives H⁺ ions); the others are bases (give OH⁻ ions)

Marking: ½ mark for HNO₃, ½ mark for the reason.

📖 Revise: Ch 3 · Acids and bases · Chapter 3 – Chemical Bonding

Raising the temperature increases the rate of a reaction mainly because

(a) the threshold energy becomes lower(b) more particles get kinetic energy equal to or greater than the threshold energy(c) the number of reactant particles increases(d) the surface area of the reactants increases

[1]

Step-by-step solution

Heating increases the kinetic energy of the particles.
So more particles reach the threshold energy (ത്രെഷോൾഡ് എനർജി) → more effective collisions → faster reaction. The threshold energy itself does not change.

✅ Answer: (b) more particles get kinetic energy equal to or greater than the threshold energy

Marking: 1 mark for (b).

📖 Revise: Ch 5 · Temperature and threshold energy · Chapter 5 – Chemical Kinetics

State whether true or false. If false, correct it: “A catalyst is used up in the reaction that it speeds up.”

[1]

Step-by-step solution

False.
A catalyst (ഉൽപ്രേരകം) increases the rate of a reaction without itself undergoing a permanent chemical change. E.g. the MnO₂ used to decompose hydrogen peroxide is recovered unchanged by filtering.

✅ Answer: False – a catalyst increases the rate without itself undergoing a permanent chemical change; it is recovered unchanged

Marking: ½ mark for “false”, ½ mark for the correct statement.

📖 Revise: Ch 5 · Catalysts · Chapter 5 – Chemical Kinetics

A torch beam is passed through four glasses in a dark room. In which glass will the path of the light be seen?

(a) salt solution(b) sugar solution(c) kanji vellam (rice water)(d) copper sulphate solution

[1]

Step-by-step solution

Salt solution, sugar solution and copper sulphate solution are true solutions – their particles (0.01–1 nm) are too small to scatter light.
Kanji vellam is a colloid (a sol). Its particles (1–1000 nm) scatter light, so the path is seen – the Tyndall effect (ടിൻഡൽ പ്രഭാവം).

✅ Answer: (c) kanji vellam (rice water)

Marking: 1 mark for (c).

📖 Revise: Ch 6 · Colloids and the Tyndall effect · Chapter 6 – Solutions

The main component of natural gas, used as CNG in autorickshaws and buses, is

(a) butane(b) methane(c) propane(d) ethene

[1]

Step-by-step solution

Natural gas (പ്രകൃതിവാതകം) is mainly methane, CH₄.
CNG is methane compressed at high pressure; butane is the main component of LPG.

✅ Answer: (b) methane

Marking: 1 mark for (b).

📖 Revise: Ch 8 · Natural gas, CNG and LNG · Chapter 8 – Organic Chemistry

10.

Find the relation and fill in: kerosene : C₁₀–C₁₆ :: diesel :

[1]

Step-by-step solution

Table 8.9 gives the number of carbon atoms in each fraction of petroleum.
Kerosene: C₁₀–C₁₆ (domestic fuel); diesel: C₁₆–C₂₀ (fuel for diesel engines).

✅ Answer: C₁₆–C₂₀

Marking: 1 mark.

📖 Revise: Ch 8 · Fractions of petroleum – Table 8.9 · Chapter 8 – Organic Chemistry

Section BAnswer all questions · 8 × 2 = 16 marks

11.

J. J. Thomson found the charge-to-mass ratio (e/m) of the electron to be 1.76 × 10¹¹ C/kg. Robert Millikan’s oil-drop experiment gave the charge of the electron as 1.6 × 10⁻¹⁹ C.
(a) Find the mass of an electron.
(b) About how many electrons, taken together, have the same mass as one proton? Why is the mass of the electrons not counted in the mass number?

[2]

Step-by-step solution

Given: e/m = 1.76 × 10¹¹ C/kg; charge e = 1.6 × 10⁻¹⁹ C
Formula: mass m = e ÷ (e/m)
Substitution: m = 1.6 × 10⁻¹⁹ ÷ (1.76 × 10¹¹) = (1.6 ÷ 1.76) × 10^{−19 − 11} = 0.909 × 10⁻³⁰ kg
Answer: mass of an electron ≈ 9.1 × 10⁻³¹ kg
(b) The mass of an electron is 1/1837 of the mass of a proton, so about 1837 electrons have the mass of one proton.
An electron’s mass is only 0.000548 u – practically 0. So the mass number counts only the protons and neutrons (each about 1 u).

✅ Answer: (a) 9.1 × 10⁻³¹ kg (b) about 1837 electrons; an electron’s mass (0.000548 u) is so small that it is taken as 0

Marking: (a) formula ½, answer with unit ½. (b) 1837 ½, reason ½.

📖 Revise: Ch 1 · Mass of the electron · Ch 1 · Table 1.2 · Chapter 1 – Structure of the Atom

12.

Look at this list of compounds: ferrous sulphate crystals (FeSO₄·7H₂O), sodium chloride (NaCl), potassium dichromate (K₂Cr₂O₇), magnesium chloride (MgCl₂), cobalt nitrate (Co(NO₃)₂), calcium carbonate (CaCO₃).
(a) Which of them would you expect to be coloured? Give the reason.
(b) In which groups are the transition elements placed, and from which period do they begin?

[2]

Step-by-step solution

(a) Compounds of transition elements are generally coloured (നിറമുള്ള സംയുക്തങ്ങൾ). Fe, Cr and Co are transition elements.
So ferrous sulphate (pale green), potassium dichromate (orange) and cobalt nitrate (pink) are coloured. NaCl, MgCl₂ and CaCO₃ have only main group elements – they are white.
(b) Transition elements (സംക്രമണ മൂലകങ്ങൾ) are in groups 3 to 12. They begin in period 4 (scandium, Z = 21).

✅ Answer: (a) FeSO₄·7H₂O, K₂Cr₂O₇ and Co(NO₃)₂ – they contain transition elements (Fe, Cr, Co) (b) groups 3–12, from period 4

Marking: (a) the three compounds ½ (any two of three still ½), reason ½. (b) groups 3–12 ½, period 4 ½.

📖 Revise: Ch 2 · Transition elements and coloured compounds · Chapter 2 – Periodic Table

13.

Milk of magnesia contains magnesium hydroxide, Mg(OH)₂. It neutralises the hydrochloric acid in the stomach.
(a) Write the ions given by magnesium hydroxide and by hydrochloric acid.
(b) Write the formula of the salt formed, showing the criss-cross steps, and write the balanced equation for the reaction.

[2]

Step-by-step solution

(a) Magnesium hydroxide (a base) gives Mg²⁺ and OH⁻ ions; hydrochloric acid gives H⁺ and Cl⁻ ions.
(b) The salt takes the positive ion of the base and the negative ion of the acid: Mg²⁺ and Cl⁻.
Write Mg Cl → swap the charge numbers (2 and 1) as subscripts → Mg₁Cl₂ → omit 1 → MgCl₂ (magnesium chloride). Check: (+2) + 2(−1) = 0.
Neutralisation (നിർവീരീകരണ രാസപ്രവർത്തനം): Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O (Mg 1 = 1, O 2 = 2, H 4 = 4, Cl 2 = 2).

✅ Answer: (a) Mg²⁺ and OH⁻; H⁺ and Cl⁻ (b) MgCl₂; Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Marking: (a) ½ (all four ions). (b) formula with steps 1, balanced equation ½.

📖 Revise: Ch 3 · Writing formulae of salts · Chapter 3 – Chemical Bonding

14.

Find the oxidation number of (a) chlorine in potassium chlorate, KClO₃ (b) nitrogen in sodium nitrite, NaNO₂. Show the working.

[2]

Step-by-step solution

Given: oxidation numbers K = +1, Na = +1, O = −2 (Table 4.7)
Rule: the sum of the oxidation numbers of all the atoms in a compound is 0.
(a) KClO₃: (+1) + x + 3(−2) = 0 → x − 5 = 0 → x = +5
(b) NaNO₂: (+1) + x + 2(−2) = 0 → x − 3 = 0 → x = +3
Answer: oxidation number (ഓക്സിഡേഷൻ നമ്പർ) of Cl in KClO₃ = +5; of N in NaNO₂ = +3.

✅ Answer: (a) +5 (b) +3

Marking: Each part: equation ½, answer ½.

📖 Revise: Ch 4 · Finding oxidation numbers · Chapter 4 – Redox Reactions

15.

In a teacher demonstration, a little solid zinc sulphate is mixed with solid barium nitrate in a dry test tube. Nothing seems to happen. When solutions of the two are mixed, a white precipitate forms at once.
(a) Name the factor that affects the rate here. Why do the solutions react at once but the solids do not?
(b) Complete and balance: ZnSO₄ + Ba(NO₃)₂ → ______ + ______ . Name the type of the reaction.

[2]

Step-by-step solution

(a) Factor: the physical state (ഭൗതികാവസ്ഥ) of the reactants.
In the solids the particles are held in fixed places and touch only at the surfaces of the grains, so there are very few collisions. In solution the ions move freely and collide very often → many effective collisions → the reaction is fast.
(b) The ions exchange partners: ZnSO₄ + Ba(NO₃)₂ → BaSO₄↓ + Zn(NO₃)₂ (already balanced). The white precipitate is barium sulphate.
Two compounds exchange ions to give two new compounds → double decomposition (ദ്വിവിഘടന രാസപ്രവർത്തനം) – a precipitation reaction.

✅ Answer: (a) physical state – in solution the particles move freely and collide much more often (b) ZnSO₄ + Ba(NO₃)₂ → BaSO₄↓ + Zn(NO₃)₂; double decomposition (precipitation)

Marking: (a) factor ½, reason ½. (b) products ½, type ½.

📖 Revise: Ch 5 · Physical state and rate · Ch 5 · Double decomposition · Chapter 5 – Chemical Kinetics

16.

Write why each of these is added to food or drinks: (a) sodium benzoate (b) allyl hexanoate (c) indigo carmine (d) vanillin

[2]

Step-by-step solution

Table 6.5 (food additives):
(a) Sodium benzoate – a preservative (പ്രിസർവേറ്റീവുകൾ) in sauces, salads and soft drinks; it acts against yeast and moulds.
(b) Allyl hexanoate – gives a pineapple flavour.
(c) Indigo carmine – gives a blue colour.
(d) Vanillin – gives flavour (vanilla).

✅ Answer: (a) preservative against yeast and moulds (b) pineapple flavour (c) blue colour (d) flavour (vanilla)

Marking: ½ mark each.

📖 Revise: Ch 6 · Additives – Table 6.5 · Chapter 6 – Solutions

17.

Two bottles P and Q contain white powders: one is potassium chloride and the other potassium carbonate. Solutions of both are made.
(a) When barium chloride solution is added, Q gives a white precipitate but P does not. Dilute hydrochloric acid is then added to the precipitate. What will you see, and what is Q?
(b) How will you confirm that P is a chloride? Write the equation.

[2]

Step-by-step solution

(a) Carbonate test (Table 7.3): K₂CO₃ + BaCl₂ → BaCO₃↓ + 2KCl (white precipitate).
On adding dilute HCl the precipitate dissolves with effervescence – carbon dioxide is given off: BaCO₃ + 2HCl → BaCl₂ + H₂O + CO₂↑. So Q is potassium carbonate (കാർബണേറ്റുകൾ).
(b) Add silver nitrate solution to P → a curdy white precipitate of silver chloride: KCl + AgNO₃ → AgCl↓ + KNO₃.
Add ammonium hydroxide (NH₄OH) → the precipitate dissolves. This confirms that P is a chloride.

✅ Answer: (a) the precipitate dissolves with effervescence (CO₂); Q is potassium carbonate (b) AgNO₃ gives a curdy white precipitate that dissolves in NH₄OH; KCl + AgNO₃ → AgCl↓ + KNO₃

Marking: (a) observation ½, Q identified ½. (b) test with both observations ½, equation ½.

📖 Revise: Ch 7 · Test for carbonates · Ch 7 · Test for chlorides · Chapter 7 – Non-metals

18.

Petrol (C₅–C₁₀) is a mixture of hydrocarbons. Its calorific value in Fig 7.2 is given as that of octane, the alkane with 8 carbon atoms.
(a) Using the general formula, find the number of hydrogen atoms in a molecule of octane. Write its molecular formula.
(b) Find the number of hydrogen atoms in decene, the 10-carbon alkene. Write its molecular formula.

[2]

Step-by-step solution

Given: octane – alkane, n = 8; decene – alkene, n = 10
Formulae: alkanes C_nH_2n+2; alkenes C_nH_2n
(a) H = 2n + 2 = 2 × 8 + 2 = 18 → octane is C₈H₁₈
(b) H = 2n = 2 × 10 = 20 → decene is C₁₀H₂₀

✅ Answer: (a) 18 – C₈H₁₈ (b) 20 – C₁₀H₂₀

Marking: Each part: number ½, formula ½.

📖 Revise: Ch 8 · General formula of alkanes · Ch 8 · General formula of alkenes · Chapter 8 – Organic Chemistry

Section CAnswer all questions · 6 × 3 = 18 marks

19.

The figure shows the orbit diagrams of three atoms X, Y and W. Only the number of neutrons is written in each nucleus.
(a) Find the atomic number and the mass number of each atom.
(b) Which two are isotopes and which two are isobars? Give the reasons.
(c) One of the isotopes is radioactive and is used to find the age of fossils; the other is used to study life processes in plants and animals. Name both.

[3]

Step-by-step solution

Given: X – electrons 2, 4 and 8 neutrons; Y – electrons 2, 5 and 7 neutrons; W – electrons 2, 4 and 7 neutrons
Formula: atomic number Z = number of electrons (= protons); mass number A = Z + number of neutrons
Substitution: X: Z = 6, A = 6 + 8 = 14 (146C); Y: Z = 7, A = 7 + 7 = 14 (147N); W: Z = 6, A = 6 + 7 = 13 (136C)
Answer: mass numbers X 14, Y 14, W 13
(b) Isotopes (ഐസോടോപ്പുകൾ): X and W – same atomic number (6), different mass numbers (14, 13).
Isobars (ഐസോബാറുകൾ): X and Y – same mass number (14), different atomic numbers (6, 7).
(c) X = carbon-14, radioactive – age of fossils. W = carbon-13 (about 1.1 % of carbon) – used to study life processes in plants and animals.

✅ Answer: (a) X: Z 6, A 14; Y: Z 7, A 14; W: Z 6, A 13 (b) isotopes X and W; isobars X and Y (c) X is carbon-14 (fossils), W is carbon-13 (life processes)

Marking: (a) all six values 1 (½ if two atoms correct). (b) isotopes ½, isobars ½ (with reasons). (c) ½ + ½.

📖 Revise: Ch 1 · Isotopes and isobars · Ch 1 · Atomic number and mass number · Chapter 1 – Structure of the Atom

20.

Nitrogen (Z = 7) combines with hydrogen (Z = 1) to form ammonia, NH₃.
(a) Draw the electron dot diagram of a molecule of ammonia.
(b) What is the valency of nitrogen in ammonia? Explain using the diagram.
(c) Using this valency of nitrogen, write the formula of calcium nitride, the compound of calcium (Z = 20) and nitrogen. Show the steps.

[3]

Step-by-step solution

(a) N: 2, 5 – five outermost electrons (dots). Each H has one electron (cross). N shares one pair with each H; one pair of N stays unshared.
N now has 8 electrons around it (octet) and each H has 2 (duplet).
(b) Nitrogen shares 3 of its electrons (three shared pairs), so its valency (സംയോജകത) in ammonia is 3.
(c) Ca: 2, 8, 8, 2 → gives 2 electrons → valency 2. N: valency 3.
Write Ca N → swap the valencies as subscripts → Ca₃N₂ (no common factor) → Ca₃N₂.

✅ Answer: (a) N with one lone pair, sharing one pair with each of three H (b) 3 – N shares three electrons (c) Ca₃N₂

Marking: (a) correct dot diagram 1 (½ if the lone pair is missing). (b) valency with reason 1. (c) valency of Ca ½, formula ½.

📖 Revise: Ch 3 · Electron dot diagrams of molecules · Ch 3 · Valency · Ch 3 · Criss-cross method · Chapter 3 – Chemical Bonding

21.

Write the oxidation number of every atom above its symbol and find out whether each is a redox reaction.
(a) 2KI + Cl₂ → 2KCl + I₂
(b) BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
(c) For the redox reaction, name the element oxidised, the element reduced, the oxidising agent and the reducing agent.

[3]

Step-by-step solution

(a) 2+1K−1I + 0Cl₂ → 2+1K−1Cl + 0I₂
I: −1 → 0 (increases); Cl: 0 → −1 (decreases); K stays +1. Oxidation numbers change → redox reaction.
(b) In Na₂SO₄: 2(+1) + x + 4(−2) = 0 → S = +6; in BaSO₄: (+2) + x + 4(−2) = 0 → S = +6.
+2Ba−1Cl₂ + +1Na₂+6S−2O₄ → +2Ba+6S−2O₄ + 2+1Na−1Cl
No oxidation number changes → not a redox reaction (it is a double decomposition).
(c) Oxidised: iodine (of KI). Reduced: chlorine. Oxidising agent (ഓക്സീകാരി): Cl₂ (it is reduced). Reducing agent (നിരോക്സീകാരി): KI (its iodide is oxidised).

✅ Answer: (a) redox: I −1 → 0, Cl 0 → −1 (b) not redox – no oxidation number changes (c) I oxidised, Cl reduced; oxidising agent Cl₂, reducing agent KI

Marking: (a) oxidation numbers ½, verdict ½. (b) oxidation numbers ½, verdict ½. (c) ½ for oxidised/reduced, ½ for the two agents.

📖 Revise: Ch 4 · Oxidation number and redox · Ch 4 · Oxidation number rules · Chapter 4 – Redox Reactions

22.

In a teacher demonstration, equal lengths (equal masses) of magnesium ribbon are dropped into two flasks of hydrochloric acid – one concentrated and one dilute. The acid is in excess both times. The hydrogen formed is collected and its volume is measured every minute: Mg + 2HCl → MgCl₂ + H₂↑. The graph shows the results.
(a) Which curve, A or B, is for the concentrated acid? Explain using collisions.
(b) From the graph, find the volume of hydrogen formed in the first minute in A and in B.
(c) In which part of each curve is the reaction fastest? Why do both curves finally become flat at the same volume?

[3]

Step-by-step solution

(a) Curve A rises faster, so it is for the concentrated acid. Higher concentration (ഗാഢത) = more acid particles per unit volume → more effective collisions with the magnesium → faster reaction.
Given (b): read the curves at t = 1 min (the volume at t = 0 is 0)
Reading: volume in the first minute = volume at 1 min − volume at 0 min
Substitution: A: 40 − 0 = 40 mL; B: 15 − 0 = 15 mL
Answer: A 40 mL, B 15 mL
(c) Both curves are steepest at the start – the most reactant is present then. The slope falls as the magnesium is used up.
The amount of product depends on the amount of reactant. Both flasks had the same mass of magnesium (the acid was in excess), so both give the same final volume, 60 mL; the concentrated acid only gives it sooner.

✅ Answer: (a) A (b) A 40 mL, B 15 mL (c) fastest at the start (steepest part); same mass of Mg gives the same 60 mL of hydrogen

Marking: (a) A ½, reason ½. (b) ½ + ½. (c) start ½, reason for the same final volume ½.

📖 Revise: Ch 5 · Concentration and rate · Ch 5 · Gas-syringe graph (extended activity) · Chapter 5 – Chemical Kinetics

23.

In a teacher demonstration in a fume cupboard, two gas jars A and B are filled with dry chlorine. A dry coloured cloth is hung in A and a moist coloured cloth of the same colour in B.
(a) In which jar does the cloth lose its colour? Why does the other cloth keep its colour?
(b) Write the equations that explain the bleaching.
(c) Is the coloured substance in the cloth oxidised or reduced? Give one use of chlorine that depends on this property.

[3]

Step-by-step solution

(a) The moist cloth in jar B loses its colour. The dry cloth in A keeps its colour, because chlorine bleaches only in the presence of water.
(b) Chlorine reacts with water to form hypochlorous acid (ഹൈപ്പോക്ലോറസ് ആസിഡ്): H₂O + Cl₂ → HCl + HOCl
Hypochlorous acid gives nascent oxygen (നവജാത ഓക്സിജൻ): HOCl → HCl + [O]
(c) The nascent oxygen oxidises the coloured substance to a colourless one – so the coloured substance is oxidised (the nascent oxygen is the oxidising agent).
Use: bleaching cloth and paper, or disinfecting drinking water (bleaching powder in wells).

✅ Answer: (a) B (moist cloth); bleaching needs water – dry chlorine gives no nascent oxygen (b) H₂O + Cl₂ → HCl + HOCl; HOCl → HCl + [O] (c) oxidised; bleaching (or disinfecting water)

Marking: (a) jar B ½, reason ½. (b) ½ + ½. (c) oxidised ½, use ½.

📖 Revise: Ch 7 · Bleaching by chlorine · Chapter 7 – Non-metals

24.

The petrol fraction of petroleum contains the alkane with 5 carbon atoms.
(a) Write its molecular formula and IUPAC name.
(b) Draw its structural formula.
(c) Balance the equation for its complete burning: C₅H₁₂ + O₂ → CO₂ + H₂O. How does burning more and more petrol add to global warming?

[3]

Step-by-step solution

(a) Alkanes: C_nH_2n+2; n = 5 → H = 2 × 5 + 2 = 12 → C₅H₁₂. Word root “pent” + “ane” → pentane.
(b) Structural formula (ഘടനാവാക്യം) – each carbon has 4 bonds, CH₃–CH₂–CH₂–CH₂–CH₃:
(c) C: 5 C on the left → 5CO₂. H: 12 H → 6H₂O. O on the right = 5 × 2 + 6 × 1 = 16 → 8O₂.
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O (C 5 = 5, H 12 = 12, O 16 = 16).
The CO₂ formed is a greenhouse gas (about 55 % share in global warming). It traps part of the infrared rays given out by the Earth, so the average temperature rises – global warming (ആഗോളതാപനം).

✅ Answer: (a) C₅H₁₂, pentane (b) five C in a chain, 12 H, every C with 4 bonds (c) C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O; CO₂ is a greenhouse gas that traps infrared rays

Marking: (a) formula ½, name ½. (b) correct structure 1. (c) balanced equation ½, global warming ½.

📖 Revise: Ch 8 · Naming alkanes · Ch 8 · Petroleum and global warming · Chapter 8 – Organic Chemistry

Section DAnswer all questions · 4 × 4 = 16 marks

25.

For the school science fair, Anu’s group made a bar chart of the approximate atomic radii of the elements of period 3 (argon, a noble gas, is left out).
(a) How does the size of the atoms change from sodium to chlorine? Explain the reason.
(b) Classify these seven elements as metals, metalloid and non-metals. Where are the metals and the non-metals placed in the periodic table?
(c) Potassium (2, 8, 8, 1) is just below sodium in group 1. Is a potassium atom larger or smaller than a sodium atom? Why?
(d) Which is larger: a potassium atom or a calcium atom (2, 8, 8, 2)? Where in the periodic table are the largest atoms found?

[4]

Step-by-step solution

(a) The size (ആറ്റത്തിന്റെ വലിപ്പം) decreases from Na (154 pm) to Cl (99 pm). All have 3 shells, but the nuclear charge (ന്യൂക്ലിയർ ചാർജ്) increases from +11 to +17. The attraction on the outermost electrons increases, so the atoms become smaller.
(b) Metals (ലോഹങ്ങൾ): Na, Mg, Al. Metalloid (ഉപലോഹങ്ങൾ): Si. Non-metals (അലോഹങ്ങൾ): P, S, Cl.
Metals are on the left (and in the middle) of the periodic table, non-metals on the right, and metalloids between them.
(c) Potassium is larger. Going down a group the nuclear charge increases, but a new shell is added (K has 4 shells, Na 3), which outweighs it – so the size increases.
(d) K and Ca are both in period 4 (4 shells). Ca has the larger nuclear charge (+20 against +19), so it pulls its electrons closer – potassium is larger.
The largest atoms are at the bottom-left of the periodic table (lower part of group 1); the smallest towards the top-right.

✅ Answer: (a) size decreases – same 3 shells, nuclear charge rises (b) metals Na, Mg, Al; metalloid Si; non-metals P, S, Cl – metals on the left, non-metals on the right (c) larger – one more shell (d) potassium; bottom-left

Marking: (a) trend ½, reason ½. (b) classification ½, positions ½. (c) larger ½, reason ½. (d) K ½, bottom-left ½.

📖 Revise: Ch 2 · Atomic size trends · Ch 2 · Metals, non-metals and metalloids · Chapter 2 – Periodic Table

26.

Coconut and arecanut farmers in Kerala spray Bordeaux mixture (copper sulphate + lime) on their palms. To check whether enough lime has been added, the farmer dips a clean iron knife into the mixture. If a reddish-brown coating forms on the knife, more lime is needed. The coating forms by this reaction:
Fe + CuSO₄ → FeSO₄ + Cu (CuSO₄ contains Cu²⁺ and SO₄²⁻ ions; FeSO₄ contains Fe²⁺ and SO₄²⁻ ions.)
(a) Using the atomic masses Fe = 56, Cu = 64, S = 32, O = 16, find the total mass of the reactants and of the products (in u). Which law does this show?
(b) Find the oxidation number of sulphur in CuSO₄ and in FeSO₄.
(c) Write the oxidation numbers of iron and copper before and after the reaction. Which is oxidised and which is reduced?
(d) Name the oxidising agent and the reducing agent. Is this a redox reaction? Why?

[4]

Step-by-step solution

Given: Fe = 56, Cu = 64, S = 32, O = 16
Formula: molecular mass = sum of the atomic masses of all the atoms
Substitution: CuSO₄ = 64 + 32 + 4 × 16 = 160 u; FeSO₄ = 56 + 32 + 4 × 16 = 152 u. Reactants: 56 + 160 = 216 u. Products: 152 + 64 = 216 u.
Answer: 216 u on both sides – the law of conservation of mass (മാസ് സംരക്ഷണനിയമം).
(b) CuSO₄: (+2) + x + 4(−2) = 0 → x = +6. FeSO₄: (+2) + x + 4(−2) = 0 → x = +6. Sulphur does not change.
(c) 0Fe + +2Cu+6S−2O₄ → +2Fe+6S−2O₄ + 0Cu
Iron: 0 → +2 – increases → iron is oxidised (Fe → Fe²⁺ + 2e⁻). Copper: +2 → 0 – decreases → copper is reduced (Cu²⁺ + 2e⁻ → Cu).
(d) Oxidising agent (ഓക്സീകാരി): CuSO₄ (Cu²⁺; it is reduced). Reducing agent (നിരോക്സീകാരി): Fe (it is oxidised).
Yes, it is a redox reaction – oxidation (of iron) and reduction (of copper) take place at the same time.

✅ Answer: (a) 216 u on both sides – law of conservation of mass (b) +6 in both (c) Fe 0 → +2 (oxidised); Cu +2 → 0 (reduced) (d) oxidising agent CuSO₄, reducing agent Fe; redox – oxidation and reduction happen together

Marking: (a) masses ½, law ½. (b) ½ + ½ (working must be shown). (c) oxidation numbers ½, oxidised/reduced ½. (d) agents ½, redox with reason ½.

📖 Revise: Ch 4 · Conservation of mass · Ch 4 · Redox reactions · Chapter 4 – Redox Reactions

27.

After the floods, a relief camp was set up in a school in Aluva.
• The water drawn from a nearby well was muddy. When it was left to stand, the mud settled; when it was filtered, mud was left on the filter paper.
• The filtered water looked clear, but when a torch beam was passed through a glass of it in a dark room, the path of the light could still be seen faintly.
• ORS solution made from packets was clear, and the path of the torch beam could not be seen in it.
• Kanji vellam (rice water) was served to the children.
(a) Is the muddy well water a true solution, a colloid or a suspension? Give two reasons.
(b) What does the faint path of light in the filtered water show? Name this effect and give the particle size of such particles.
(c) Is ORS solution a true solution, a colloid or a suspension? Write two properties of its particles.
(d) Kanji vellam is a colloid. Name its type, its dispersed phase and its dispersion medium.

[4]

Step-by-step solution

(a) Suspension (സസ്പെൻഷൻ): the particles settle when left undisturbed and are separated by filtering (particles larger than 1000 nm, often visible).
(b) Particles of colloidal size (1–1000 nm) are still present – they pass through filter paper but scatter light. This is the Tyndall effect (ടിൻഡൽ പ്രഭാവം). So the filtered water is not yet a true solution.
(c) ORS is a true solution (യഥാർഥ ലായനി). Its particles are 0.01–1 nm: they cannot be seen even with a microscope, they do not settle, they pass through filter paper, and they do not scatter light (no beam).
(d) Kanji vellam is a sol: dispersed phase (ഡിസ്പേർസ്ഡ് ഫേസ്) – solid (starch particles); dispersion medium (ഡിസ്പേർഷൻ മീഡിയം) – liquid (water).

✅ Answer: (a) suspension – particles settle and are separated by filtering (b) colloidal particles still present – Tyndall effect, 1–1000 nm (c) true solution – 0.01–1 nm, do not settle/scatter light (d) sol: solid in liquid

Marking: (a) suspension ½, two reasons ½. (b) colloidal particles ½, Tyndall effect with size ½. (c) true solution ½, two properties ½. (d) type ½, phase and medium ½.

📖 Revise: Ch 6 · True solutions, suspensions and colloids · Ch 6 · Types of colloids · Chapter 6 – Solutions

28.

Rockets designed at the Vikram Sarabhai Space Centre (VSSC), Thumba, Thiruvananthapuram, have a cryogenic stage that burns liquid hydrogen with liquid oxygen carried in the rocket.
(a) Write the balanced equation for the burning of hydrogen. Using H = 1 and O = 16, find the mass of oxygen needed to burn 2 kg of hydrogen.
(b) The calorific value of hydrogen is 143 kJ/g. How much heat is released when 2 kg of hydrogen burns completely?
(c) Oxygen itself does not burn. Why must the rocket still carry oxygen?
(d) On its way up, the rocket passes through the ozone layer in the stratosphere. How is ozone formed from oxygen there, and why is the ozone layer important?

[4]

Step-by-step solution

(a) Given: 2H₂ + O₂ → 2H₂O (H 4 = 4, O 2 = 2); H = 1, O = 16; hydrogen = 2 kg
Formula: masses react in the same ratio as in the equation: 2H₂ = 2 × 2 = 4 u; O₂ = 32 u; 2H₂O = 36 u (4 + 32 = 36 – mass is conserved)
Substitution: 4 kg of hydrogen needs 32 kg of oxygen, so 2 kg needs 32 × 2 ÷ 4 = 16 kg
Answer: 16 kg of oxygen (and 18 kg of water forms)
(b) Given: calorific value 143 kJ/g; mass = 2 kg = 2000 g
Formula: heat = mass × calorific value
Substitution: 2000 × 143
Answer: 286 000 kJ
(c) Burning needs a supporter of combustion (ജ്വലനസഹായി). Oxygen does not burn but helps hydrogen burn – it is the oxidiser of the rocket fuel. High up the air is very thin, and in space there is no air, so the rocket must carry its own oxygen.
(d) Ultraviolet rays split oxygen molecules into atoms: O₂ → 2O (UV); an oxygen atom joins an oxygen molecule: O₂ + O → O₃.
Ozone absorbs harmful ultraviolet radiations from the Sun and changes back to oxygen. This cycle keeps the ozone layer steady and protects living things from UV.

✅ Answer: (a) 2H₂ + O₂ → 2H₂O; 16 kg (b) 286 000 kJ (c) oxygen supports combustion; there is hardly any air high up (d) O₂ → 2O (UV), O₂ + O → O₃; it absorbs harmful UV

Marking: (a) balanced equation ½, 16 kg ½. (b) method ½, answer with unit ½. (c) 1 (supports combustion ½, no air high up ½). (d) formation ½, importance ½.

📖 Revise: Ch 7 · Oxygen and ozone · Ch 7 · Calorific value of hydrogen · Chapter 7 – Non-metals

Question numbers by chapter

Chapter	A (1)	B (2)	C (3)	D (4)	Total
1. Structure of the Atom	1, 2	11	19	–	7
2. Periodic Table	3	12	–	25	7
3. Chemical Bonding	4, 5	13	20	–	7
4. Redox Reactions	–	14	21	26	9
5. Chemical Kinetics	6, 7	15	22	–	7
6. Solutions	8	16	–	27	7
7. Non-metals	–	17	23	28	9
8. Organic Chemistry	9, 10	18	24	–	7
Total	10	16	18	16	60

Score to grade

Kerala grading scale: A+ 54–60 (90 %+), A 48–53, B+ 42–47, B 36–41, C+ 30–35, C 24–29, D+ 18–23, D 12–17, E below 12. For every question the student lost marks on, the linked guide page explains the idea again.