ICSE Class 10 Chemistry Sample Paper 2024 with Solutions

SECTION A

(Attempt all questions from this Section.)

Question 1
Choose the correct answers to the questions from the given options.
(Do not copy the question, write the correct answers only.) [15 Marks]

(i) An aqueous solution of copper sulphate turns colourless on electrolysis. Which of the following could be the electrodes? [1 Mark]
P. anode: copper; cathode: copper
Q. anode: platinum; cathode: copper
R. anode: copper; cathode: platinum
(a) only P
(b) only Q
(c) only R
(d) both Q and R

Answer: (b) only Q

When platinum is used as anode and copper as cathode, copper ions (\( Cu^{2+} \)) are discharged at the cathode and deposited as copper metal, while \( OH^{-} \) ions are discharged at the platinum anode releasing oxygen gas. Thus, \( Cu^{2+} \) ions are removed from the solution, turning the blue copper sulphate solution colourless.

Teacher's Note:
a) When an active electrode like a copper anode is used, copper dissolves into the solution to replace the \( Cu^{2+} \) ions discharged at the cathode, so the blue colour of the solution does not fade.
b) Students must remember that for the solution to turn colourless, the concentration of \( Cu^{2+} \) ions in the electrolyte must continuously decrease, which happens only with an inert anode.

(ii) A compound P is heated in a test tube with sodium hydroxide solution. A red litmus paper held at the mouth of the test tube turns blue. [1 Mark]
Which of the following could compound P be?
(a) zinc sulphate
(b) copper sulphate
(c) ferrous sulphate
(d) ammonium sulphate

Answer: (d) ammonium sulphate

Ammonium salts on heating with caustic alkali (NaOH) release ammonia gas, which is alkaline in nature and turns moist red litmus paper blue.

Teacher's Note:
a) Ammonium salts react with sodium hydroxide to liberate pungent-smelling ammonia gas (\( NH_3 \)).
b) Ensure students do not confuse this with metal salt solutions that precipitate metal hydroxides with NaOH without releasing an alkaline gas.

(iii) The atomic masses of sulphur (S), oxygen (O), and helium (He) are approximately 32, 16, and 4 respectively. [1 Mark]
Which of the following statements regarding the number of atoms in 32 g of sulphur, 16 g of oxygen, and 4 g of helium is correct?
P. 16 g of oxygen contains four times the number of atoms as 4 g of helium.
Q. 16 g of oxygen contains half the number of atoms as 32 g of sulphur.
(a) only P
(b) only Q
(c) both P and Q
(d) neither P nor Q

Answer: (d) neither P nor Q

Moles of S = \( 32 / 32 = 1 \) mole of atoms = \( N_A \) atoms.
Moles of O = \( 16 / 16 = 1 \) mole of molecules \( (O_2) \), which equals 2 moles of atoms = \( 2 N_A \) atoms.
Moles of He = \( 4 / 4 = 1 \) mole of atoms = \( N_A \) atoms.
Comparing atoms: 16 g of oxygen has \( 2 N_A \) atoms, 4 g of helium has \( N_A \) atoms (so 2 times, not 4 times). 32 g of sulphur has \( N_A \) atoms, whereas 16 g of oxygen has \( 2 N_A \) atoms (so twice, not half). Therefore, both statements P and Q are incorrect.

Teacher's Note:
a) Always distinguish carefully between moles of molecules and moles of atoms for gases like oxygen (\( O_2 \)).
b) A common error is assuming mass ratio directly equals atom ratio without considering atomicity.

(iv) Ammonia gas is passed through quicklime and then collected in a jar. Red and blue litmus papers are placed in the jar. W, X, Y and Z are the four observations. [1 Mark]
Which of the above observations correctly shows the reaction of the litmus papers to ammonia?

(a) W
(b) X
(c) Y
(d) Z

Answer: (a) W

Dry ammonia gas does not affect dry litmus paper, but since the litmus papers placed in the gas jar are normally moist or react with basic ammonia gas in the presence of trace moisture to form ammonium hydroxide, red litmus turns blue while blue litmus remains blue.

Teacher's Note:
a) Ammonia is a basic gas that turns red litmus blue.
b) Blue litmus shows no colour change in a basic medium, remaining blue.

(v) Glucose reacts with concentrated sulphuric acid to give a very pure form of carbon called sugar charcoal. [1 Mark]
The reaction taking place is:
(a) oxidation
(b) combustion
(c) dehydration
(d) combination

Answer: (c) dehydration

Concentrated sulphuric acid acts as a strong dehydrating agent and removes elements of water from glucose \( (C_6H_{12}O_6 \rightarrow 6C + 6H_2O) \), leaving behind black carbon (sugar charcoal).

Teacher's Note:
a) Concentrated \( H_2SO_4 \) has a high affinity for water.
b) This characteristic reaction demonstrates the dehydrating property of concentrated sulphuric acid on organic compounds containing hydrogen and oxygen in the ratio of 2:1.

(vi) In which of the following electrolytic cells [P, Q, R or S] will silver plating be done on the spoon? [1 Mark]
(a) P
(b) Q
(c) R
(d) S

[Figure: Four diagrams P, Q, R, S showing electrolytic cells with a battery, a silver plate, and a silver plated spoon. In P: anode is silver plate, cathode is silver plated spoon, electrolyte is distilled water. In Q: anode is silver plate, cathode is silver plated spoon, electrolyte is silver nitrate solution. In R: anode is silver plated spoon, cathode is silver plate, electrolyte is distilled water. In S: anode is silver plated spoon, cathode is silver plate, electrolyte is silver nitrate solution.]

Answer: (b) Q

For electroplating an article, the article to be plated must be connected to the negative terminal (cathode), the pure metal to be deposited must be connected to the positive terminal (anode), and an aqueous solution of a salt of the plating metal must be used as the electrolyte.

Teacher's Note:
a) Distilled water cannot act as an electrolyte because it lacks ions; a soluble salt solution like silver nitrate is essential.
b) The object to be electroplated must always form the cathode.

(vii) The basicity of acetic acid is: [1 Mark]
(a) 1
(b) 2
(c) 3
(d) 4

Answer: (a) 1

Acetic acid \( (CH_3COOH) \) contains one replaceable hydrogen ion per molecule in its carboxyl group, making its basicity equal to 1.

Teacher's Note:
a) Basicity refers to the number of ionizable hydrogen atoms present in one molecule of an acid.
b) Organic acids like acetic acid are monobasic despite having four hydrogen atoms because only one hydrogen is attached to oxygen and ionizes.

(viii) A \(\rightarrow\) A⁺³; B \(\rightarrow\) B^-1 [1 Mark]
Number of electrons present in the outermost shell of atoms A and B respectively are:
(a) 5, 1
(b) 3, 1
(c) 3, 7
(d) 5, 7

Answer: (c) 3, 7

Atom A loses 3 electrons to form \( \text{A}^{+3} \), meaning it has 3 valence electrons. Atom B gains 1 electron to form \( \text{B}^{-1} \), meaning it has 7 valence electrons.

Teacher's Note:
a) Metals lose valence electrons to form positive ions equal in magnitude to the number of electrons lost.
b) Non-metals gain electrons to complete their octet, and the number of electrons gained corresponds to the deficiency from 8.

(ix) A __________ solution is observed after placing Magnesium metal in a solution of Copper sulphate for half an hour. [1 Mark]
(a) Blue
(b) Colourless
(c) Reddish brown
(d) Dirty green

Answer: (b) Colourless

Magnesium being more reactive than copper displaces copper from copper sulphate solution, forming colourless magnesium sulphate solution and reddish-brown copper metal.

Teacher's Note:
a) The blue colour of copper sulphate fades as blue \( Cu^{2+} \) ions are replaced by colourless \( Mg^{2+} \) ions.
b) Students should note that while zinc turns copper sulphate solution colourless, iron turns it light green, and magnesium turns it colourless.

(x) An element with atomic no. __________ will form an acidic oxide. [1 Mark]
(a) 3
(b) 17
(c) 11
(d) 13

Answer: (b) 17

Element with atomic number 17 is Chlorine (electronic configuration 2, 8, 7), which is a non-metal. Non-metals form acidic or neutral oxides.

Teacher's Note:
a) Atomic numbers 3 (Lithium) and 11 (Sodium) are alkali metals forming basic oxides, 13 (Aluminium) is amphoteric, and 17 (Chlorine) is a non-metal forming an acidic oxide.
b) Metallic oxides are generally basic, whereas non-metallic oxides are acidic.

(xi) Which of the following is NOT true with respect to nitric acid? [1 Mark]
(a) It is a strong reducing agent
(b) It is a strong oxidizing agent
(c) It is unstable to heat
(d) It liberates sulphur dioxide gas when treated with potassium sulphite

Answer: (a) It is a strong reducing agent

Nitric acid is a very strong oxidizing agent, not a reducing agent. Statements (b), (c), and (d) are true properties of nitric acid.

Teacher's Note:
a) Nitric acid readily decomposes to release nascent oxygen, making it a powerful oxidizing agent.
b) It reacts with metal sulphites like potassium sulphite to liberate sulphur dioxide gas.

(xii) __________ is the functional group in methanol. [1 Mark]
(a) >C=O
(b) –OH
(c) –CHO
(d) –COOH

Answer: (b) –OH

Methanol is an alcohol containing the hydroxyl functional group (–OH).

Teacher's Note:
a) Alcohols are characterized by the hydroxyl (–OH) functional group.
b) >C=O is carbonyl (ketone), –CHO is aldehyde, and –COOH is carboxyl (carboxylic acid).

(xiii) The process of electrolysis is an example of: [1 Mark]
(a) Oxidation reaction
(b) Reduction reaction
(c) Redox reaction
(d) Displacement reaction

Answer: (c) Redox reaction

During electrolysis, oxidation occurs at the anode (loss of electrons) and reduction occurs at the cathode (gain of electrons), making it a redox reaction.

Teacher's Note:
a) Electrolysis involves both simultaneous oxidation and reduction processes.
b) Anions are oxidized at the anode and cations are reduced at the cathode.

(xiv) The catalyst used in Ostwald's process is ___________. [1 Mark]
(a) Finely divided iron
(b) Graphite
(c) Vanadium pentoxide
(d) Platinum

Answer: (d) Platinum

Platinum gauze is used as a catalyst in the catalytic oxidation of ammonia to nitric oxide in the Ostwald's process.

Teacher's Note:
a) Finely divided iron is used in Haber's process, and vanadium pentoxide is used in Contact process.
b) Platinum is the specific catalyst for Ostwald's process maintained at about \( 800^{\circ}C \).

(xv) An element belongs to third period and sixteenth group. It will have __________ electrons in its valence shell. [1 Mark]
(a) 2
(b) 5
(c) 6
(d) 3

Answer: (c) 6

For elements in groups 13 to 18, the number of valence electrons is equal to Group Number - 10. Thus, Group 16 elements have \( 16 - 10 = 6 \) valence electrons.

Teacher's Note:
a) The group number in the modern periodic table for groups 13-18 directly indicates valence electrons when 10 is subtracted.
b) The element is Sulphur, which has electronic configuration 2, 8, 6.

Question 2

(i) The setup shown below is that of the fountain experiment with hydrogen chloride gas in the flask. [5 Marks]
[Figure: Diagram of fountain experiment showing a round-bottom flask filled with HCl gas, a jet tube passing through a rubber bung, a dropper containing water, and a beaker containing litmus solution. The setup is mounted on a stand.]
The fountain starts when a few drops of water from the dropper are introduced into the flask. Instead of the drops of water, Pooja started the fountain by introducing a few drops of Sodium hydroxide into the flask.
(a) Explain why the litmus solution gets sucked up when Sodium hydroxide is used.
(b) What will be the colour of the fountain when Sodium hydroxide is used? Justify your answer.
(c) If instead of HCl gas, ammonia gas is filled in the flask and water is introduced from the dropper, will there be a different observation? Justify your answer.

Answer:
(a) Sodium hydroxide reacts rapidly with hydrogen chloride gas to form sodium chloride and water. This causes a sudden and drastic decrease in pressure inside the flask, creating a partial vacuum, which forces the external litmus solution to rush up into the flask through the jet tube.
(b) The colour of the fountain will be blue. Sodium hydroxide is a strong alkali, and when mixed with neutral or red litmus solution, it turns the solution blue.
(c) No, there will be no different observation regarding the formation of the fountain; a fountain will still be formed because ammonia gas is extremely soluble in water. However, the colour of the fountain will be blue instead of red because aqueous ammonia is alkaline in nature.

Teacher's Note:
a) The fountain experiment demonstrates the high solubility of gases like HCl and \( NH_3 \) in water.
b) Emphasize that the creation of the partial vacuum due to high solubility or rapid chemical absorption is the core principle behind the fountain.

(ii) Match the following Column A with Column B. [5 Marks]

Answer:
(a) Aluminium - 3. Hall Heroult's process
(b) Sulphuric acid - 4. Contact Process
(c) Calcination - 2. Carbonate ore
(d) Calcium Chloride - 5. Electrovalent compound
(e) Carbon tetrachloride - 1. Covalent compound

Teacher's Note:
a) Match each term based on industrial extraction methods, chemical processes, ore types, and bonding types.
b) Verify that ionic compounds are formed by electron transfer and covalent compounds by electron sharing.

(iii) Complete the following by choosing the correct answers from the bracket: [5 Marks]
(a) If an element has one electron in the outermost shell then it is likely to have the ___________ [smallest/ largest] atomic size amongst all the elements in the same period.
(b) ___________ [sulphuric acid/ hydrochloric acid] does not form an acid salt.
(c) A __________ [reddish brown/ dirty green] coloured precipitate is formed when ammonium hydroxide is added to a solution of ferric chloride.
(d) Alkanes undergo __________ [addition/ substitution] reactions.
(e) An __________ [alkaline/acidic] solution will turn methyl orange solution pink.

Answer:
(a) largest
(b) hydrochloric acid
(c) reddish brown
(d) substitution
(e) acidic

Teacher's Note:
a) Atomic size decreases across a period from left to right; therefore, group 1 elements with one valence electron have the largest size.
b) Hydrochloric acid is monobasic and forms only normal salts, whereas sulphuric acid is dibasic and can form both normal and acid salts.

(iv) Identify the following: [5 Marks]
(a) A bond formed between two atoms by sharing of a pair of electrons, with both electrons being provided by the same atom.
(b) A salt formed by the complete neutralization of an acid by a base.
(c) A reaction in which the hydrogen of an alkane is replaced by a halogen.
(d) The energy required to remove an electron from a neutral gaseous atom.
(e) A homogenous mixture of two or more metals or a metal and a non-metal in a definite proportion in their molten state.

Answer:
(a) Coordinate bond (or Dative bond)
(b) Normal salt
(c) Substitution reaction
(d) Ionization potential (or Ionization energy)
(e) Alloy

Teacher's Note:
a) Ensure precise chemical terminology is used for definitions.
b) Distinguish clearly between normal salts formed by complete neutralization and acid salts formed by partial neutralization.

(v) (a) Draw the structural diagram for the following compounds: [5 Marks]
1. 1-propanal
2. 1, 2-dichloro ethane
3. But-2-ene
(b) Give the IUPAC name of the following organic compounds:
1.
[Figure: Structural formula showing \( H-C(H)(H)-C(H)(H)-C(OH)(H)-H \)]
2.
[Figure: Structural formula showing \( H-C(H)(H)-C(H)=C(H)-C(H)(H)-C(H)(H)-H \)]

Answer:
(a) Structural diagrams:
1. 1-propanal: \( CH_3 - CH_2 - CHO \) (Three carbon chain with terminal aldehyde group –CHO, middle \( -CH_2- \), and terminal methyl \( -CH_3 \)).
2. 1,2-dichloro ethane: \( Cl-CH_2 - CH_2-Cl \) (Two carbons bonded together, each bonded to two hydrogens and one chlorine atom).
3. But-2-ene: \( CH_3 - CH = CH - CH_3 \) (Four carbons in a chain with a double bond between carbon 2 and carbon 3).
(b) IUPAC names:
1. Propan-1-ol (or 1-propanol)
2. Pent-2-ene

Teacher's Note:
a) In drawing structures, ensure all covalent bonds (lines representing shared electron pairs) for carbon, hydrogen, chlorine, and oxygen are correctly shown.
b) IUPAC naming requires numbering the longest carbon chain from the end that gives the lowest locant to the functional group or double bond.

SECTION B

(Attempt any four questions.)

Question 3

(i) Identify the reactant and write the balanced equation for the following: [2 Marks]
Nitric acid reacts with compound Q to give a salt \( Ca(NO_3)_2 \), water and carbon dioxide.

Answer:
Reactant Q is Calcium carbonate \( (CaCO_3) \) [or Calcium bicarbonate \( Ca(HCO_3)_2 \)].
Balanced equation:
\( CaCO_3 + 2HNO_3 \rightarrow Ca(NO_3)_2 + H_2O + CO_2 \uparrow \)

Teacher's Note:
a) Metal carbonates react with dilute acids to form salt, water, and carbon dioxide gas with effervescence.
b) Verify balancing by checking that both sides have equal numbers of calcium, nitrogen, hydrogen, carbon, and oxygen atoms.

(ii) What property of Sulphuric acid is exhibited in each of the following cases: [2 Marks]
(a) In the preparation of HCl gas when it reacts with Sodium chloride.
(b) When conc. Sulphuric acid reacts with Copper to produce Sulphur dioxide gas.

Answer:
(a) Non-volatile nature (or high boiling point)
(b) Oxidizing property (strong oxidizing agent)

Teacher's Note:
a) Non-volatile concentrated sulphuric acid displaces more volatile hydrochloric acid from its salt.
b) Concentrated sulphuric acid oxidizes copper to copper sulphate while getting reduced itself to sulphur dioxide.

(iii) The electron affinity of an element X is greater than that of element Y. [3 Marks]
(a) How is the oxidising power of X likely to compare with that of Y?
(b) How is the electronegativity of X likely to compare with that of Y?
(c) State whether X is likely to be placed to the left or to the right of Y in the periodic table?

Answer:
(a) The oxidizing power of X is greater than that of Y.
(b) The electronegativity of X is greater than that of Y.
(c) X is likely to be placed to the right of Y in the periodic table.

Teacher's Note:
a) Electron affinity, electronegativity, and oxidizing power all increase across a period from left to right.
b) Higher electron affinity means a stronger tendency to accept electrons, which corresponds to stronger oxidizing behaviour.

(iv) (a) State whether the following statements are TRUE or FALSE. Justify your answer. [3 Marks]
1. In an electrovalent compound, the cation attains the electronic configuration of the noble gas that comes after it in the periodic table.
2. In the formation of a compound \( PQ_2 \), atom P gives one electron to each atom of Q. The compound \( PQ_2 \) is good conductor of electricity.
(b) Calculate the number of moles in 22 grams of carbon dioxide.

Answer:
(a) 1. FALSE. The cation is formed by loss of electrons from a metal atom, so it attains the electronic configuration of the noble gas that comes before it in the periodic table.
2. FALSE. If atom P gives one electron to each of two Q atoms, P loses 2 electrons (valency 2) and each Q gains 1 electron. While it is an electrovalent compound, solid ionic compounds do not conduct electricity; they conduct electricity only in molten or aqueous state.
(b) Molar mass of \( CO_2 = 12 + (16 \times 2) = 44 \text{ g/mol} \).
Number of moles = \( \text{Mass} / \text{Molar Mass} = 22 / 44 = 0.5 \text{ moles} \).

Teacher's Note:
a) Cations (metals like Na, atomic number 11) lose an electron to attain the nearest preceding noble gas configuration (Neon, atomic number 10).
b) Ionic compounds conduct electricity solely in molten or aqueous solutions due to free mobile ions, not in solid state.

Question 4

(i) The following questions relate to the extraction of Aluminium by electrolysis. [2 Marks]
(a) Name the other aluminum containing compound added to alumina.
(b) Give a balanced equation for the reaction that takes place at the cathode.

Answer:
(a) Cryolite \( (Na_3AlF_6) \) [or Fluorspar \( CaF_2 \)]
(b) Cathode reaction: \( Al^{3+} + 3e^{-} \rightarrow Al \)

Teacher's Note:
a) Cryolite acts as a solvent for pure alumina and lowers its fusion temperature while increasing electrical conductivity.
b) Aluminium ions migrate to the cathode and undergo reduction by gaining electrons to form molten aluminium metal.

(ii) A gas cylinder of capacity \( 40 \text{ dm}^3 \) is filled with gas X the mass of which is 20 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of hydrogen is 2 g. Find the relative molecular mass of the gas. [2 Marks]

Answer:
According to Avogadro's Law, equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.
Mass of \( 40 \text{ dm}^3 \) of gas X = 20 g.
Mass of \( 40 \text{ dm}^3 \) of Hydrogen gas = 2 g.
Vapour density of gas X = \( \text{Mass of volume V of gas X} / \text{Mass of same volume of hydrogen} = 20 / 2 = 10 \).
Relative Molecular Mass = \( 2 \times \text{Vapour Density} = 2 \times 10 = 20 \).

Teacher's Note:
a) Use the ratio of masses of equal volumes of gas and hydrogen to calculate vapour density directly.
b) Recall the fundamental relation: Molecular Mass = 2 \(\times\) Vapour Density.

(iii) Give balanced equations for each of the following: [3 Marks]
(a) Action of warm water on Aluminium nitride.
(b) Oxidation of carbon with conc. Nitric acid.
(c) Dehydration of ethanol by conc. Sulphuric acid at a temperature of \( 170^{\circ}C \).

Answer:
(a) \( AlN + 3H_2O \rightarrow Al(OH)_3 \downarrow + NH_3 \uparrow \)
(b) \( C + 4HNO_3\text{(conc.)} \rightarrow CO_2 + 2H_2O + 4NO_2 \uparrow \)
(c) \( C_2H_5OH \xrightarrow[170^{\circ}C]{\text{conc. } H_2SO_4} C_2H_4 \uparrow + H_2O \)

Teacher's Note:
a) Ensure all state symbols or evolution/precipitation arrows are correctly indicated where applicable.
b) Note that carbon is oxidized to carbon dioxide by concentrated nitric acid while nitric acid is reduced to nitrogen dioxide.

(iv) With respect to Haber's process answer the following: [3 Marks]
(a) Temperature of the reaction
(b) Catalyst used
(c) Balanced equation for the reaction occurring

Answer:
(a) Temperature: \( 450^{\circ}C \) to \( 500^{\circ}C \)
(b) Catalyst: Finely divided iron (promoter: Molybdenum)
(c) Balanced equation:
\( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) (in the presence of Fe catalyst, Mo promoter, and high pressure of 200 - 900 atm)

Teacher's Note:
a) Haber's process is a reversible exothermic reaction used for the industrial manufacture of ammonia.
b) Students must mention the optimum temperature, pressure, and catalyst conditions precisely.

Question 5

(i) (a) Ranjana wants to prove that ammonia is a reducing agent. To demonstrate this, she passes ammonia gas over heated copper oxide. What will she observe? [2 Marks]
(b) Write a balanced chemical equation for the above reaction.

Answer:
(a) Observation: The black copper oxide turns reddish-brown due to the formation of copper metal, and droplets of a colourless liquid (water) appear on the cooler parts of the test tube.
(b) Balanced equation:
\( 3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2 \uparrow \)

Teacher's Note:
a) Ammonia reduces metal oxides to respective metals by losing hydrogen and getting oxidized to nitrogen gas.
b) The colour change from black to reddish-brown confirms reduction of copper(II) oxide.

(ii) Name the alloy which is made up of: [2 Marks]
(a) Copper, Zinc and Tin
(b) Lead and Tin

Answer:
(a) Brass (Note: Brass is Cu and Zn; Bronze is Cu and Sn. Copper, Zinc and Tin together form Gunmetal. Let us write Gunmetal for Cu, Zn and Sn).
(b) Solder

Teacher's Note:
a) Gunmetal contains copper, zinc, and tin.
b) Solder is an alloy of lead and tin used for joining metals due to its low melting point.

(iii) Seema takes a blue crystalline salt P in a test tube. On heating it produces a white anhydrous powder. P is dissolved in water. Zinc is added to one part of the solution and to another part of the solution Barium chloride is added. [3 Marks]
(a) Name the compound P.
(b) Mention one observation when zinc is added to the solution of P.
(c) State the colour of the precipitate formed when barium chloride is added to the solution of P.

Answer:
(a) Compound P is Hydrated Copper Sulphate [Copper(II) sulphate pentahydrate, \( CuSO_4 \cdot 5H_2O \)].
(b) Observation: The blue colour of the solution gradually fades and reddish-brown copper metal gets deposited on the zinc granules.
(c) Colour of precipitate: White precipitate (of Barium sulphate, \( BaSO_4 \)).

Teacher's Note:
a) Hydrated copper sulphate loses its water of crystallization on heating to form white anhydrous copper sulphate.
b) Barium chloride reacts with sulphate solutions to give a white precipitate insoluble in dilute HCl.

(iv) Give reasons: [3 Marks]
(a) Ethene undergoes addition reaction.
(b) Hydrocarbons can be used as fuels.
(c) Hydrogen chloride gas cannot be collected over water.

Answer:
(a) Ethene is an unsaturated hydrocarbon containing a carbon-carbon double bond (\( C=C \)), which readily breaks to allow atoms to add across the double bond.
(b) Hydrocarbons burn in air or oxygen with the evolution of a large amount of heat energy, making them excellent exothermic fuels.
(c) Hydrogen chloride gas is extremely soluble in water, dissolving instantly to form hydrochloric acid rather than displacing water.

Teacher's Note:
a) Unsaturated compounds are chemically reactive due to the presence of pi bonds in multiple bonds.
b) Highly soluble gases like HCl and \( NH_3 \) must be collected by upward displacement of air or downward delivery.

Question 6

(i) Name the following: [2 Marks]
(a) The ore of Zinc containing its sulphide.
(b) The most commonly used oxide ore of Aluminium.

Answer:
(a) Zinc blende (Sphalerite, \( ZnS \))
(b) Bauxite (\( Al_2O_3 \cdot 2H_2O \))

Teacher's Note:
a) Zinc blende is concentrated by froth flotation process.
b) Bauxite is purified by Bayer's process to obtain pure alumina.

(ii) State one observation in the following cases: [2 Marks]
(a) Sodium chloride solution is added to a solution of lead nitrate.
(b) Barium chloride solution is added to a solution of Zinc sulphate.

Answer:
(a) A white precipitate of lead chloride is formed.
(b) A white precipitate of barium sulphate is formed.

Teacher's Note:
a) Lead chloride precipitates as a white solid which is soluble in hot water.
b) Barium sulphate is a white insoluble precipitate formed due to the reaction of barium ions with sulphate ions.

(iii) Copper sulphate solution is electrolysed using copper electrodes. [3 Marks]
(a) Which electrode [cathode or anode] is the oxidizing electrode? Why?
(b) Write the equation for the reaction occurring at the above electrode.

Answer:
(a) Anode is the oxidizing electrode because oxidation (loss of electrons) takes place at the anode.
(b) Equation at the anode:
\( Cu - 2e^{-} \rightarrow Cu^{2+} \) (or \( Cu \rightarrow Cu^{2+} + 2e^{-} \)

Teacher's Note:
a) Oxidation is defined as loss of electrons, which occurs at the positive electrode (anode) during electrolysis.
b) Copper anode dissolves into the electrolyte during this process.

(iv) X [2, 8, 7] and Y [2, 8, 2] are two elements. Using this information complete the following: [3 Marks]
(a) __________ is the metallic element.
(b) Metal atoms tend to have a maximum of __________ electrons in the outermost shell.
(c) ___________ is the reducing agent.

Answer:
(a) Y
(b) 3 (or 1, 2, or 3)
(c) Y

Teacher's Note:
a) Element Y with configuration 2, 8, 2 has 2 valence electrons, making it a metal (Magnesium), whereas X with 7 valence electrons is a non-metal.
b) Metals act as reducing agents as they readily lose electrons.

Question 7

(i) The empirical formula of an organic compound is \( C_3H_4N \). Its molecular weight is 108. Find the amount of carbon in one mole of the compound. Show all the steps involved. (Atomic weights: C- 12; H- 1; N- 14) [3 Marks]

Answer:
1. Empirical formula mass of \( C_3H_4N = (3 \times 12) + (4 \times 1) + 14 = 36 + 4 + 14 = 54 \text{ g/mol} \).
2. n = Molecular weight / Empirical formula mass = \( 108 / 54 = 2 \).
3. Molecular formula = \( (C_3H_4N)_n = (C_3H_4N)_2 = C_6H_8N_2 \).
4. Amount of carbon in one mole of the compound = \( 6 \times \text{Atomic weight of C} = 6 \times 12 = 72 \text{ grams} \).

Teacher's Note:
a) First determine the value of n by dividing the molecular weight by the empirical formula mass.
b) Multiply the subscripts in the empirical formula by n to find the molecular formula before calculating total mass of carbon.

(ii) (a) Mahesh prepared a basic solution X that has a pH 7. [3 Marks]
How will the pH of the solution X change on addition of the following:
1. Hydrochloric acid
2. a solution of a base
(b) The atomic number of an element is 15. To which group will this element belong to?

Answer:
(a) 1. On adding hydrochloric acid, the pH of the solution will decrease (become less than 7, towards acidic range).
2. On adding a solution of a base, the pH of the solution will increase (become greater than 7, towards alkaline range).
(b) Electronic configuration of element with atomic number 15 is 2, 8, 5. Since it has 3 shells, it belongs to Period 3, and since it has 5 valence electrons, its Group is \( 5 + 10 = 15 \).

Teacher's Note:
a) pH scale ranges from 0 to 14, where values below 7 are acidic, 7 is neutral, and above 7 are basic.
b) Group number for elements in groups 13-18 is calculated as 10 + number of valence electrons.

(iii) 8.2 grams of calcium nitrate is decomposed by heating according to the equation: [4 Marks]
\( 2Ca(NO_3)_2 \rightarrow 2CaO + 4NO_2 + O_2 \)
Calculate the following:
(a) Volume of nitrogen dioxide obtained at STP
(b) Mass of CaO formed
[Atomic weights: Ca = 40, N = 14, O = 16]

Answer:
1. Molecular mass of \( Ca(NO_3)_2 = 40 + 2 \times (14 + 16 \times 3) = 40 + 2 \times 62 = 40 + 124 = 164 \text{ g/mol} \).
Mass of 2 moles of \( Ca(NO_3)_2 = 2 \times 164 = 328 \text{ g} \).
2. According to the balanced equation, 328 g of \( Ca(NO_3)_2 \) produces 4 moles of \( NO_2 \) (which is \( 4 \times 22.4 \text{ litres} = 89.6 \text{ litres} \) at STP) and \( 2 \) moles of \( CaO \) (mass = \( 2 \times (40 + 16) = 2 \times 56 = 112 \text{ g} \)).
(a) Volume of \( NO_2 \) obtained from 8.2 g of \( Ca(NO_3)_2 \):
Volume = \( (89.6 / 328) \times 8.2 = 2.24 \text{ litres} \).
(b) Mass of \( CaO \) obtained from 8.2 g of \( Ca(NO_3)_2 \):
Mass = \( (112 / 328) \times 8.2 = 2.8 \text{ grams} \).

Teacher's Note:
a) Always use stoichiometry from the balanced chemical equation based on molar masses and molar volumes.
b) 1 mole of any gas at STP occupies 22.4 litres.

Question 8

(i) State giving reasons if: [2 Marks]
(a) zinc and aluminium can be distinguished by heating the metal powder with concentrated sodium hydroxide solution.
(b) calcium nitrate and lead nitrate can be distinguished by adding ammonium hydroxide solution to the salt solution.

Answer:
(a) No, zinc and aluminium cannot be distinguished using concentrated NaOH solution because both metals react with hot concentrated sodium hydroxide to liberate hydrogen gas and form soluble zincate and aluminate respectively.
(b) Yes, they can be distinguished. Lead nitrate gives a chalky white precipitate of lead hydroxide which is insoluble in excess of \( NH_4OH \), whereas calcium nitrate does not form any precipitate with \( NH_4OH \) because calcium hydroxide is sparingly soluble and \( Ca^{2+} \) is not precipitated by ammonium hydroxide.

Teacher's Note:
a) Both zinc and aluminium are amphoteric metals reacting with alkalis.
b) Analytical chemistry tests using sodium hydroxide or ammonium hydroxide provide confirmatory precipitates for identifying specific cations.

(ii) Draw the electron dot diagram of Hydronium ion. [2 Marks]

Answer:
The hydronium ion (\( H_3O^{+} \)) is formed by a coordinate covalent bond between a water molecule (\( H_2O \)) and a hydrogen ion (\( H^{+} \)).
[Figure: Electron dot diagram showing oxygen atom surrounded by 8 electrons (two O-H single covalent bonds with hydrogen atoms, one coordinate bond sharing a lone pair of electrons with \( H^{+} \), and one unshared lone pair of electrons on oxygen), enclosed in square brackets with a positive charge \( + \) outside the bracket.]

Answer:
Oxygen atom contributes 6 valence electrons and each of the two hydrogen atoms contributes 1 electron, forming a water molecule with two lone pairs on oxygen. When a proton (\( H^{+} \) with no electrons) approaches, oxygen donates one of its lone pairs to form a coordinate bond, resulting in \( H_3O^{+} \) with an overall positive charge.

Teacher's Note:
a) Clearly show the lone pairs on oxygen and the coordinate bond arrow or shared electron pair.
b) Enclose the entire electron dot structure in square brackets with a superscript \( + \) to represent the cation.

(iii) Give balanced equations for the following: [3 Marks]
(a) Laboratory preparation of ethyne from calcium carbide.
(b) Conversion of acetic acid to ethyl acetate.
(c) Laboratory preparation of nitric acid.

Answer:
(a) \( CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 \uparrow \)
(b) \( CH_3COOH + C_2H_5OH \xrightarrow[\text{conc. } H_2SO_4}{\text{heat}} CH_3COOC_2H_5 + H_2O \)
(c) \( NaNO_3 + H_2SO_4\text{(conc.)} \xrightarrow[< 200^{\circ}C]{} NaHSO_4 + HNO_3 \)

Teacher's Note:
a) Calcium carbide reacts with cold water to produce ethyne (acetylene) gas.
b) The esterification reaction between acetic acid and ethanol requires concentrated sulphuric acid as a dehydrating and catalytic agent.

(iv) Identify the following substances: [3 Marks]
(a) An alkaline gas which produces dense white fumes when reacted with HCl gas.
(b) The anion present in the salt, which produces a gas with the smell of rotten eggs when reacted with dil. HCl.
(c) The particles present in strong electrolytes.

Answer:
(a) Ammonia gas (\( NH_3 \))
(b) Sulphide anion (\( S^{2-} \))
(c) Free mobile ions

Answer:
(a) Ammonia gas reacts with hydrogen chloride to form ammonium chloride, which appears as dense white fumes.
(b) Metal sulphides react with dilute HCl to release hydrogen sulphide gas (\( H_2S \)), which has the characteristic smell of rotten eggs.
(c) Strong electrolytes dissociate almost completely in solution to yield a high concentration of free mobile ions.

Teacher's Note:
a) Memorize characteristic gas tests (smell, colour, precipitate formation, and fume reactions).
b) Strong electrolytes conduct electricity efficiently due to the presence of 100% dissociated mobile ions.