Class 9 · ChemistrySet 1 · Answer key

Answer Key – Set 1 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 1, 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.

📄 Question paper💻 Practise onlineAll sets
ANSWER KEY & MARKING SCHEME · SET 1
Standard IX · Chemistry · Annual Examination
Total score: 60 · Time: 2 hours
Section AAnswer all questions · 10 × 1 = 10 marks
1.
In Rutherford’s gold foil experiment, most of the α particles passed straight through the gold foil without being deflected. From this observation Rutherford concluded that
(a) the nucleus is positively charged(b) most of the space inside an atom is empty(c) electrons revolve only in fixed orbits(d) the atom is a positive sphere with electrons embedded in it
[1]

Step-by-step solution

  1. If the α particles met nothing on their way, there must be nothing in most of the atom to stop or turn them. So most of the atom is empty space (ഗോൾഡ് ഫോയിൽ പരീക്ഷണം).
  2. The positive nucleus (a) was concluded from the few particles that were deflected or bounced back; (c) is Bohr’s idea; (d) is Thomson’s plum pudding model.
✅ Answer: (b) most of the space inside an atom is empty

Marking: 1 mark for (b).

📖 Revise: Ch 1 · Gold foil experiment · Chapter 1 – Structure of the Atom

2.
When a magnesium atom (2, 8, 2) changes into a magnesium ion, which of these does not change?
(a) the number of electrons(b) the number of protons(c) the charge(d) the electron configuration
[1]

Step-by-step solution

  1. The atom gives away its 2 outermost electrons: Mg → Mg2+ + 2e−.
  2. The number of electrons (12 → 10), the charge (0 → +2) and the configuration (2, 8, 2 → 2, 8) all change. Only electrons move; the nucleus is untouched, so the number of protons stays 12.
✅ Answer: (b) the number of protons

Marking: 1 mark for (b).

📖 Revise: Ch 3 · Table 3.6 – Mg before and after · Chapter 3 – Chemical Bonding

3.
The energy needed to remove the most loosely bound electron from the outermost shell of an isolated gaseous atom is called .
[1]

Step-by-step solution

  1. This is the definition of ionisation energy (അയോണീകരണ ഊർജം), also called ionisation enthalpy.
✅ Answer: ionisation energy (ionisation enthalpy)

Marking: 1 mark.

📖 Revise: Ch 3 · Ionisation energy · Chapter 3 – Chemical Bonding

4.
In the reaction 4Na + O2 → 2Na2O, the substance that is reduced is
(a) sodium(b) oxygen(c) sodium oxide(d) both sodium and oxygen
[1]

Step-by-step solution

  1. Sodium atoms lose electrons: Na → Na+ + e− (oxidation number 0 → +1). Sodium is oxidised.
  2. Oxygen atoms gain electrons: O + 2e− → O2− (0 → −2). Gain of electrons is reduction (നിരോക്സീകരണം), so oxygen is reduced.
✅ Answer: (b) oxygen

Marking: 1 mark for (b).

📖 Revise: Ch 4 · Oxidising and reducing agents · Chapter 4 – Redox Reactions

5.
A candle is burnt on an open electronic balance. The reading slowly decreases. This is because
(a) mass is destroyed when the candle burns(b) the gases formed – carbon dioxide and water vapour – escape into the air(c) the melted wax becomes lighter(d) the law of conservation of mass does not apply to burning
[1]

Step-by-step solution

  1. By the law of conservation of mass (മാസ് സംരക്ഷണനിയമം) the total mass of the products equals that of the reactants.
  2. The products of burning are gases that escape from the open pan, so the balance does not weigh them. If they were collected and weighed, no mass would be missing.
✅ Answer: (b) the gases formed – carbon dioxide and water vapour – escape into the air

Marking: 1 mark for (b).

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

6.
In the Haber process for making ammonia, N2(g) + 3H2(g) ⇌ 2NH3(g), the catalyst used is
(a) manganese dioxide(b) vanadium pentoxide(c) iron(d) amylase
[1]

Step-by-step solution

  1. In the Haber process iron (Fe) is the catalyst (ഉൽപ്രേരകം), at about 200 atm and 700 K.
  2. Vanadium pentoxide is the catalyst in the contact process; manganese dioxide speeds up the decomposition of hydrogen peroxide; amylase is an enzyme in saliva.
✅ Answer: (c) iron

Marking: 1 mark for (c).

📖 Revise: Ch 5 · Catalysts – Haber and contact processes · Chapter 5 – Chemical Kinetics

7.
Enamel paint on a brush does not wash off with water, but it comes off easily in turpentine. The reason is
(a) turpentine is heavier than water(b) like dissolves like – non-polar paint dissolves in the non-polar solvent turpentine(c) water is a non-polar solvent(d) the paint reacts chemically with turpentine
[1]

Step-by-step solution

  1. Enamel paint is non-polar. Water is a polar solvent; turpentine is non-polar.
  2. A solute dissolves in a solvent of the same kind: like dissolves like (സമാനമായത് സമാനമായതിനെ ലയിപ്പിക്കും).
✅ Answer: (b) like dissolves like – non-polar paint dissolves in the non-polar solvent turpentine

Marking: 1 mark for (b).

📖 Revise: Ch 6 · Organic solvents, like dissolves like · Chapter 6 – Solutions

8.
A 916 gold ornament has 916 g of gold in every 1000 g; the rest is silver and copper. In this solid solution, the solvent is .
[1]

Step-by-step solution

  1. The component present in the larger amount is the solvent (ലായകം).
  2. Gold is 916 g out of 1000 g, far more than silver and copper together (84 g), so gold is the solvent; silver and copper are the solutes.
✅ Answer: gold

Marking: 1 mark.

📖 Revise: Ch 6 · 916 gold box · Chapter 6 – Solutions

9.
Which of these is a saturated hydrocarbon?
(a) C3H4(b) C4H10(c) C2H2(d) C2H4
[1]

Step-by-step solution

  1. Saturated hydrocarbons are the alkanes, with only single bonds; their general formula is CnH2n+2.
  2. C4H10: n = 4, 2 × 4 + 2 = 10 ✔ – butane, an alkane. C3H4 and C2H2 fit CnH2n−2 (alkynes) and C2H4 fits CnH2n (an alkene) – all unsaturated.
✅ Answer: (b) C4H10

Marking: 1 mark for (b).

📖 Revise: Ch 8 · Alkanes – C<sub>n</sub>H<sub>2n+2</sub> · Chapter 8 – Organic Chemistry

10.
Find the relation and fill in:   CH3–CH3 : ethane  ::  CH2=CH–CH3 :
[1]

Step-by-step solution

  1. CH2=CH–CH3 has 3 carbon atoms → word root prop.
  2. It has a double bond → suffix -ene. Name: propene.
✅ Answer: propene

Marking: 1 mark.

📖 Revise: Ch 8 · Word roots and suffixes · Chapter 8 – Organic Chemistry

Section BAnswer all questions · 8 × 2 = 16 marks
11.
An atom of an element X has mass number 19 and has 10 neutrons in its nucleus. (a) Find its atomic number and the number of electrons in it. (b) Write its electron configuration. How many electrons are there in its outermost shell?
[2]

Step-by-step solution

  1. Given: mass number A = 19, neutrons = 10
  2. Formula: A = number of protons + number of neutrons, so atomic number Z (അറ്റോമിക നമ്പർ) = A − neutrons
  3. Substitution: Z = 19 − 10
  4. Answer: Z = 9. In a neutral atom, electrons = protons = 9.
  5. (b) Fill K first (2), then L: 9 = 2 + 7 → configuration (ഇലക്ട്രോൺ വിന്യാസം) 2, 7. The outermost shell (L) has 7 electrons. (X is fluorine.)
✅ Answer: (a) atomic number 9, electrons 9   (b) 2, 7 – 7 outermost electrons

Marking: Atomic number ½, electrons ½, configuration ½, outermost electrons ½.

📖 Revise: Ch 1 · A − Z · Ch 1 · Electron configuration · Chapter 1 – Structure of the Atom

12.
(a) State Mendeleev’s periodic law and the modern periodic law. (b) Chlorine has two isotopes, 35
17Cl and 37
17Cl. Why would they be a problem in a table arranged by atomic mass, but not in the modern periodic table?
[2]

Step-by-step solution

  1. (a) Mendeleev’s periodic law (പീരിയോഡിക് നിയമം, 1869): the properties of elements are periodic functions of their atomic masses.
  2. Modern periodic law (ആധുനിക പീരിയോഡിക് നിയമം, Moseley): the chemical and physical properties of elements are periodic functions of their atomic numbers.
  3. (b) The two isotopes have different mass numbers (35 and 37), so by atomic mass they would need two different places – Mendeleev’s table had no separate places for isotopes.
  4. Both have the same atomic number, 17. The modern table is arranged by atomic number, so both share the one place of chlorine.
✅ Answer: (a) properties are periodic functions of atomic masses (Mendeleev) / of atomic numbers (modern) (b) different atomic masses but the same atomic number

Marking: (a) each law ½; (b) reason 1 (½ for the problem by atomic mass, ½ for the same atomic number).

📖 Revise: Ch 2 · Mendeleev – merits and limitations · Ch 2 · Modern periodic law · Chapter 2 – Periodic Table

13.
Calcium (Z = 20) and sulphur (Z = 16) combine to form the ionic compound calcium sulphide. Complete the table for the ions in it.
IonNumber of protonsNumber of electronsElectron configuration
Ca2+
S2−
[2]

Step-by-step solution

  1. Ca 2, 8, 8, 2 gives away its 2 outermost electrons → Ca2+ (a cation, കാറ്റയോൺ).
  2. S 2, 8, 6 takes 2 electrons → S2− (an anion, ആനയോൺ). Protons never change.
  3. IonProtonsElectronsConfiguration
    Ca2+2020 − 2 = 182, 8, 8
    S2−1616 + 2 = 182, 8, 8
  4. Both ions have the octet configuration 2, 8, 8 – the same as the noble gas argon.
✅ Answer: Ca2+: 20 protons, 18 electrons, 2, 8, 8   S2−: 16 protons, 18 electrons, 2, 8, 8

Marking: Ca2+ row 1 (protons and electrons ½, configuration ½); S2− row 1 (same split).

📖 Revise: Ch 3 · Ions and their configurations · Ch 3 · Octet · Chapter 3 – Chemical Bonding

14.
Aluminium powder burns in oxygen:   4Al + 3O2 → 2Al2O3. Using the atomic masses Al = 27 u and O = 16 u, find the total mass of the reactants and the total mass of the products. Which law does the result show?
[2]

Step-by-step solution

  1. Given: Al = 27 u, O = 16 u; 4 Al atoms, 3 O2 molecules → 2 Al2O3
  2. Formula: total mass = sum of (number of particles × mass of one particle)
  3. Substitution: reactants = 4 × 27 + 3 × (2 × 16) = 108 + 96; products = 2 × (2 × 27 + 3 × 16) = 2 × (54 + 48) = 2 × 102
  4. Answer: reactants = 204 u, products = 204 u
  5. The masses are equal: the law of conservation of mass (മാസ് സംരക്ഷണനിയമം) – atoms are only rearranged.
✅ Answer: reactants 204 u, products 204 u – the law of conservation of mass

Marking: Mass of reactants with working 1, mass of products with working ½, law ½.

📖 Revise: Ch 4 · Tables 4.1–4.2 · Chapter 4 – Redox Reactions

15.
A, B, C and D stand for atoms or ions. Name the type of chemical reaction shown by each pattern.
(i) A + B → AB    (ii) AB → A + B    (iii) A + BC → AC + B    (iv) AB + CD → AD + CB
[2]

Step-by-step solution

  1. (i) Two substances join to give one product: combination (സംയോജന രാസപ്രവർത്തനം), e.g. CaO + H2O → Ca(OH)2.
  2. (ii) One compound breaks into two substances: decomposition (വിഘടന രാസപ്രവർത്തനം), e.g. CaCO3 → CaO + CO2↑.
  3. (iii) Element A takes the place of B in compound BC: displacement (ആദേശ രാസപ്രവർത്തനം), e.g. Zn + 2HCl → ZnCl2 + H2↑.
  4. (iv) Two compounds exchange their ions: double decomposition (ദ്വിവിഘടന രാസപ്രവർത്തനം), e.g. NaCl + AgNO3 → AgCl↓ + NaNO3.
✅ Answer: (i) combination (ii) decomposition (iii) displacement (iv) double decomposition

Marking: ½ mark for each.

📖 Revise: Ch 5 · Combination and decomposition · Ch 5 · Double decomposition and displacement · Chapter 5 – Chemical Kinetics

16.
For pickling lemons, Ammini’s mother needs 250 g of a salt solution that is 8 % salt by mass. How much salt and how much water must she take?
[2]

Step-by-step solution

  1. Given: mass of solution = 250 g, mass percentage = 8 %
  2. Formula: mass percentage (മാസ് പെർസെന്റേജ്) = mass of solute ÷ mass of solution × 100, so mass of solute = mass % × mass of solution ÷ 100
  3. Substitution: mass of salt = 8 × 250 ÷ 100; mass of water = 250 − 20
  4. Answer: salt = 20 g, water = 230 g
✅ Answer: 20 g of salt and 230 g of water

Marking: Formula ½, mass of salt 1, mass of water ½.

📖 Revise: Ch 6 · Mass percentage · Chapter 6 – Solutions

17.
Raju wants to show that washing soda (അലക്കുകാരം, Na2CO3·10H2O) is a carbonate. (a) He adds barium chloride solution to a solution of washing soda. What does he see? (b) He then adds dilute hydrochloric acid to what was formed. What does he see, which gas is given off, and how can this gas be identified?
[2]

Step-by-step solution

  1. (a) A white precipitate of barium carbonate forms: Na2CO3 + BaCl2 → BaCO3↓ + 2NaCl.
  2. (b) The precipitate dissolves with effervescence (bubbles): BaCO3 + 2HCl → BaCl2 + H2O + CO2↑.
  3. The gas is carbon dioxide. Passed into lime water (ചുണ്ണാമ്പുവെള്ളം) it turns the lime water milky. So the salt is a carbonate.
✅ Answer: (a) a white precipitate (b) it dissolves with effervescence; carbon dioxide; it turns lime water milky

Marking: (a) white precipitate ½; (b) dissolves with effervescence ½, carbon dioxide ½, lime water turns milky ½.

📖 Revise: Ch 7 · Table 7.3 – test for carbonates · Chapter 7 – Non-metals

18.
(a) How many hydrogen atoms are there in a molecule of the alkane that has 12 carbon atoms? (b) An alkene has 16 hydrogen atoms in its molecule. How many carbon atoms does it have? Write its molecular formula.
[2]

Step-by-step solution

  1. Given: alkane, n = 12; alkene, number of H = 16
  2. Formula: general formula (പൊതുവാക്യം) of alkanes CnH2n+2; of alkenes CnH2n
  3. Substitution: (a) H = 2 × 12 + 2; (b) 2n = 16, so n = 16 ÷ 2
  4. Answer: (a) 26 hydrogen atoms (C12H26); (b) 8 carbon atoms – C8H16
✅ Answer: (a) 26   (b) 8 – C8H16

Marking: (a) formula ½, answer ½; (b) working ½, answer with formula ½.

📖 Revise: Ch 8 · C<sub>n</sub>H<sub>2n+2</sub> · Ch 8 · C<sub>n</sub>H<sub>2n</sub> · Chapter 8 – Organic Chemistry

Section CAnswer all questions · 6 × 3 = 18 marks
19.
The atomic numbers of three elements P, Q and R are 5, 12 and 16 (the letters are not the real symbols). Write the electron configuration of each, and find its period number and group number. Give the reasons.
[3]

Step-by-step solution

  1. Rules: period number = number of shells. Group 1 and 2: group number = number of outermost electrons. Groups 13–18: group number = outermost electrons + 10 (the ten groups 3–12 come in between).
  2. P (Z = 5): 2, 3 → 2 shells → period 2; 3 outermost electrons → 3 + 10 = group 13.
  3. Q (Z = 12): 2, 8, 2 → 3 shells → period 3; 2 outermost electrons → group 2.
  4. R (Z = 16): 2, 8, 6 → 3 shells → period 3; 6 outermost electrons → 6 + 10 = group 16.
  5. (P is boron, Q is magnesium, R is sulphur.)
✅ Answer: P 2, 3 – period 2, group 13 · Q 2, 8, 2 – period 3, group 2 · R 2, 8, 6 – period 3, group 16

Marking: For each element: configuration ½, period and group with the reason ½ (3 × 1).

📖 Revise: Ch 2 · Finding the group and period number · Chapter 2 – Periodic Table

20.
Potassium (Z = 19) burns in oxygen (Z = 8) to form the ionic compound potassium oxide. (a) Write the electron configurations of potassium and oxygen. How many electrons does each potassium atom give away, and how many does an oxygen atom take? (b) Draw the electron dot diagram to show the formation of potassium oxide. (c) Write the electron configurations of the two ions and the formula of the compound.
[3]

Step-by-step solution

  1. (a) K: 2, 8, 8, 1 – it gives away its 1 outermost electron to reach an octet. O: 2, 6 – it takes 2 electrons to reach an octet.
  2. One oxygen atom needs 2 electrons but one potassium atom gives only 1, so two potassium atoms are needed.
  3. (b) Electron dot diagram (ഇലക്ട്രോൺ ഡോട്ട് ഡയഗ്രം) – only outermost electrons are drawn; × = electrons of potassium, • = electrons of oxygen:K×+O+K×+K2−O××+K(2, 8, 8)(2, 8)(2, 8, 8)
  4. (c) K+: 2, 8, 8 (like argon); O2−: 2, 8 (like neon). Formula: K2O.
  5. The oppositely charged ions are held by electrostatic attraction – an ionic bond (അയോണിക ബന്ധനം). (Equation: 4K + O2 → 2K2O.)
✅ Answer: (a) K 2, 8, 8, 1 gives 1; O 2, 6 takes 2 (b) diagram below (c) K+ 2, 8, 8; O2− 2, 8; K2O

Marking: (a) configurations and electrons lost/gained 1; (b) correct dot diagram with brackets and charges 1; (c) ion configurations ½, formula ½.

📖 Revise: Ch 3 · Electron dot diagrams · Ch 3 · Ionic compounds · Chapter 3 – Chemical Bonding

21.
Balance the following equations.
(a) Al + HCl → AlCl3 + H2
(b) C2H4 + O2 → CO2 + H2O  (burning of ethene)
(c) Fe2O3 + H2 → Fe + H2O  (hydrogen as a reducing agent)
[3]

Step-by-step solution

  1. Change only the numbers in front of the formulae (coefficients), never the subscripts. Count each kind of atom on both sides.
  2. (a) Cl: 3 in AlCl3, so put 6 in front of HCl for two AlCl3 → 2Al; then 6 H on the left → 3H2. 2Al + 6HCl → 2AlCl3 + 3H2
    AtomAlHCl
    Left side266
    Right side266
  3. (b) 2 C → 2CO2; 4 H → 2H2O; O on the right = 4 + 2 = 6 → 3O2. C2H4 + 3O2 → 2CO2 + 2H2O
    AtomCHO
    Left side246
    Right side246
  4. (c) 2 Fe on the left → 2Fe; 3 O → 3H2O; then 6 H → 3H2. Fe2O3 + 3H2 → 2Fe + 3H2O
    AtomFeOH
    Left side236
    Right side236
✅ Answer: (a) 2Al + 6HCl → 2AlCl3 + 3H2   (b) C2H4 + 3O2 → 2CO2 + 2H2O   (c) Fe2O3 + 3H2 → 2Fe + 3H2O

Marking: 1 mark for each balanced equation (½ if the atom count is shown but one coefficient is wrong).

📖 Revise: Ch 4 · Balancing equations · Chapter 4 – Redox Reactions

22.
At a juice shop, sugar syrup is made by dissolving 60 g of sugar in 240 g of water. (a) Find the mass percentage of sugar in the syrup. (b) For one glass of kulukki sarbath, 50 g of this syrup is mixed with 150 g of water. How many grams of sugar are there in the glass? (c) Find the mass percentage of sugar in the sarbath. (Ignore the lime juice and basil seeds.)
[3]

Step-by-step solution

  1. Given: sugar (solute, ലീനം) 60 g, water (solvent, ലായകം) 240 g
  2. Formula: mass percentage = mass of solute ÷ mass of solution × 100; mass of solution = solute + solvent
  3. Substitution: (a) 60 ÷ (60 + 240) × 100 = 60 ÷ 300 × 100
  4. Answer: (a) 20 %
  5. (b) 20 % means 20 g of sugar in every 100 g of syrup, so 50 g of syrup has 20 × 50 ÷ 100 = 10 g of sugar.
  6. (c) Mass of the drink = 50 + 150 = 200 g; mass percentage = 10 ÷ 200 × 100 = 5 %.
✅ Answer: (a) 20 % (b) 10 g (c) 5 %

Marking: (a) formula ½, answer ½; (b) 1; (c) mass of the drink ½, answer ½.

📖 Revise: Ch 6 · Mass percentage · Chapter 6 – Solutions

23.
The figure shows how hydrogen is made in the laboratory. (a) Name A (a metal) and B (an acid). (b) Write the balanced equation for the reaction and name the type of reaction. (c) What is observed when the burning splinter C is brought to the mouth of the test tube? What does it show?C – burning splinterAB
[3]

Step-by-step solution

  1. (a) A – zinc (Zn) pieces; B – dilute hydrochloric acid (HCl).
  2. (b) Zn + 2HCl → ZnCl2 + H2↑. Zinc takes the place of hydrogen in HCl, so it is a displacement reaction (ആദേശ രാസപ്രവർത്തനം).
  3. (c) The gas burns with a “pop” sound (പോപ്പ് ശബ്ദം). This shows the gas is hydrogen.
✅ Answer: (a) A zinc, B dilute hydrochloric acid (b) Zn + 2HCl → ZnCl₂ + H₂ – displacement (c) the gas burns with a “pop” sound – it is hydrogen

Marking: (a) ½ + ½; (b) balanced equation ½, type ½; (c) observation ½, conclusion ½.

📖 Revise: Ch 7 · Laboratory preparation of hydrogen · Chapter 7 – Non-metals

24.
The carbon skeletons of three hydrocarbons P, Q and R are given (hydrogen atoms not shown). (a) Complete each structure by adding hydrogen atoms so that every carbon atom has four bonds. (b) Write the molecular formula of each and say whether it is an alkane, alkene or alkyne. (c) Write the IUPAC name of each.PCCCCCQCCCCRCCC
[3]

Step-by-step solution

  1. Carbon has valency 4. For each carbon, H atoms = 4 − (number of bonds to other carbons; a double bond counts 2, a triple bond 3).
  2. P: PCHHHCHHCHHCHHCHHHcondensed: CH3–CH2–CH2–CH2–CH3 → C5H12, alkane → pentane
  3. Q: QCHHCHCHHCHHHcondensed: CH2=CH–CH2–CH3 → C4H8, alkene → butene
  4. R: RCHCCHHHcondensed: CH≡C–CH3 → C3H4, alkyne → propyne
  5. Check with the general formulas: pentane 2 × 5 + 2 = 12 H; butene 2 × 4 = 8 H; propyne 2 × 3 − 2 = 4 H.
✅ Answer: P: C5H12, alkane, pentane · Q: C4H8, alkene, butene · R: C3H4, alkyne, propyne

Marking: For each compound: structure with every carbon having 4 bonds ½, molecular formula, family and name ½ (3 × 1).

📖 Revise: Ch 8 · Alkenes and alkynes · Ch 8 · IUPAC names · Chapter 8 – Organic Chemistry

Section DAnswer all questions · 4 × 4 = 16 marks
25.
Read the situation and answer. On a visit to the old house of her grandmother, Meera found a CRT television – a cathode ray tube. Her teacher showed the class this picture of the discharge tube used by scientists such as William Crookes.
(a) Name the electrodes P and Q. (b) When the tube contains air at normal pressure, no current flows through it even at a very high voltage. Why? What is done with R to make the discharge take place? (c) The glow at the ZnS coating stays the same whatever gas is taken in the tube and whatever metal the electrodes are made of. What did scientists conclude from this? Name the particles. (d) Goldstein used a cathode with holes in it in a similar tube and found rays coming out behind the cathode. Name these rays and give one way in which they differ from the rays from P.to vacuum pumphigh voltage−+PQRZnS coating
[4]

Step-by-step solution

  1. (a) P is joined to the negative terminal – the cathode (കാഥോഡ്). Q is joined to the positive terminal – the anode, here a perforated anode (സുഷിരങ്ങളുള്ള ആനോഡ്).
  2. (b) Air at normal pressure is an insulator (വൈദ്യുതരോധി), so no current flows. The air is pumped out through R with a vacuum pump (നിർവാത പമ്പ്) step by step until the pressure is very low; then the electric discharge takes place.
  3. (c) The cathode rays (കാഥോഡ് രശ്മികൾ) do not depend on the gas or the metal, so the particles in them must be present in all matter. These particles are electrons.
  4. (d) Canal rays (കനാൽ രശ്മികൾ), also called anode rays. Difference (any one): canal rays are positively charged while cathode rays are negative; the nature of canal rays depends on the gas (hydrogen gives the lightest particles – protons), but cathode rays do not.
✅ Answer: (a) P cathode, Q anode (perforated) (b) air at normal pressure is an insulator; the air is pumped out through R to a very low pressure (c) the particles are present in all matter – electrons (d) canal (anode) rays; they are positively charged / their nature depends on the gas

Marking: (a) ½ + ½; (b) insulator ½, reduce the pressure through R ½; (c) present in all matter ½, electrons ½; (d) name ½, difference ½.

📖 Revise: Ch 1 · Discharge tube and cathode rays · Ch 1 · Canal rays · Chapter 1 – Structure of the Atom

26.
Sea shells burnt to make lime (chunnambu) in Kerala contain the element X. The orbit diagram of an atom of X is given.
(a) Write the atomic number, the mass number and the electron configuration of X. (b) Find the period number and the group number of X. Give the reasons. (c) Name the family of X. Name one other element (Z ≤ 20) with similar chemical properties and say why they are similar. (d) Element Y has one proton fewer than X. Write the configuration of Y. Are X and Y in the same period or in the same group?20p20nX
[4]

Step-by-step solution

  1. (a) Protons = 20 → atomic number 20. Mass number = 20 + 20 = 40. Electrons in the shells: 2, 8, 8, 2. (X is calcium.)
  2. (b) 4 shells → period 4 (period number = number of shells). 2 outermost electrons → group 2 (for groups 1 and 2, group number = number of outermost electrons).
  3. (c) Group 2 is the family of alkaline earth metals (ആൽക്കലൈൻ എർത്ത് ലോഹങ്ങൾ). Magnesium (2, 8, 2) or beryllium (2, 2) – they also have 2 outermost electrons (ബാഹ്യതമ ഇലക്ട്രോണുകൾ), and chemical properties depend on the outermost electrons.
  4. (d) Y has 19 protons → 19 electrons → 2, 8, 8, 1 (potassium). Y also has 4 shells, so X and Y are in the same period (period 4); Y is in group 1, X in group 2.
✅ Answer: (a) Z = 20, A = 40, 2, 8, 8, 2 (b) period 4, group 2 (c) alkaline earth metals; magnesium (or beryllium) – same number (2) of outermost electrons (d) 2, 8, 8, 1; the same period (4)

Marking: (a) Z and A ½, configuration ½; (b) period with reason ½, group with reason ½; (c) family ½, element with reason ½; (d) configuration ½, same period ½.

📖 Revise: Ch 2 · Group and period number · Ch 2 · Families · Chapter 2 – Periodic Table

27.
Read the passage and answer. Raghavan, a coconut farmer, sprays Bordeaux mixture on his palms to protect them from മഹാളി (mahali). Bordeaux mixture is made from copper sulphate solution and lime. The lime is made by strongly heating sea shells (calcium carbonate) to get quicklime, and then adding water to the quicklime, which becomes very hot. The agricultural officer warns him not to keep copper sulphate solution in an iron bucket.
(a) CaCO3 → CaO + CO2↑ – name the type of this reaction. (b) CaO + H2O → Ca(OH)2 – name the type of reaction and write one sign that shows a chemical change. (c) Write the equation for the reaction of iron with copper sulphate solution. Name its type and say what Raghavan would see in the bucket. (d) When the two are mixed: CuSO4 + Ca(OH)2 → Cu(OH)2↓ + CaSO4. Name the type of this reaction and give the reason.
[4]

Step-by-step solution

  1. (a) One compound breaks into two substances: decomposition – heating is used, so it is a thermal decomposition (താപീയ വിഘടനം).
  2. (b) Two substances combine to give one compound: combination. Sign: the mixture becomes very hot – heat is given out (an exothermic reaction, താപമോചക പ്രവർത്തനം).
  3. (c) Fe + CuSO4 → FeSO4 + Cu – iron replaces copper in the compound: displacement. A reddish-brown coat of copper appears on the iron and the blue colour of the solution fades. (The bucket is eaten away and the copper sulphate is wasted.)
  4. (d) Double decomposition (ദ്വിവിഘടന രാസപ്രവർത്തനം): the two compounds exchange their ions – Cu2+ joins OH−, and Ca2+ joins SO42−, forming two new compounds.
✅ Answer: (a) decomposition (b) combination; it becomes very hot (c) Fe + CuSO₄ → FeSO₄ + Cu – displacement; a reddish-brown coat of copper forms and the blue colour fades (d) double decomposition – the two compounds exchange ions

Marking: Each part 1 mark: (a) type 1; (b) type ½, sign ½; (c) equation and type ½, observation ½; (d) type ½, reason ½.

📖 Revise: Ch 5 · Combination and decomposition · Ch 5 · Displacement and double decomposition · Chapter 5 – Chemical Kinetics

28.
A school science club in Kochi is comparing fuels for a project on cleaner transport. The bar chart (textbook Fig 7.2) gives the calorific values of some fuels.
(a) What is meant by calorific value? Which fuel in the chart has the highest calorific value? (b) Find the heat released when 5 g of hydrogen burns completely. (c) What mass of carbon monoxide must burn completely to give the same amount of heat? (d) Write one merit and one limitation of hydrogen as a fuel.hydrogen143methane56petrol (octane)48coal (carbon)33ethanol30methanol23carbohydrates16carbon monoxide10050100150calorific value (kJ/g)
[4]

Step-by-step solution

  1. (a) Calorific value (കലോറിക മൂല്യം) is the heat energy released when unit mass of a fuel burns completely (kJ/g). Hydrogen has the highest value, 143 kJ/g.
  2. Given: hydrogen 143 kJ/g, carbon monoxide 10 kJ/g; mass of hydrogen 5 g
  3. Formula: heat released = mass × calorific value; mass = heat ÷ calorific value
  4. Substitution: (b) heat = 5 × 143; (c) mass of CO = 715 ÷ 10
  5. Answer: (b) 715 kJ; (c) 71.5 g of carbon monoxide – more than 14 times the mass of hydrogen.
  6. (d) Merit: very high calorific value, and burning it gives only water (2H2 + O2 → 2H2O), so there is little pollution. Limitation: it burns explosively, and it is difficult to store and transport.
✅ Answer: (a) heat energy released when unit mass of a fuel burns completely; hydrogen (b) 715 kJ (c) 71.5 g (d) merit: high calorific value / only water forms; limitation: burns explosively / hard to store and transport

Marking: (a) definition ½, hydrogen ½; (b) 1; (c) 1; (d) merit ½, limitation ½.

📖 Revise: Ch 7 · Calorific value – Fig 7.2 · Chapter 7 – Non-metals

Question numbers by chapter

ChapterA (1)B (2)C (3)D (4)Total
1. Structure of the Atom111–257
2. Periodic Table–1219269
3. Chemical Bonding2, 31320–7
4. Redox Reactions4, 51421–7
5. Chemical Kinetics615–277
6. Solutions7, 81622–7
7. Non-metals–1723289
8. Organic Chemistry9, 101824–7
Total1016181660

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.