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ICSE Class X Prelims 2025 : Chemistry (Karnataka ICSE Schools Association KISA, Bengaluru) Answers with Marking Scheme

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KARNATAKA ICSE SCHOOLS ASSOCIATION ICSE STD. X Preparatory Examination 2025 Subject Chemistry Duration: 2 Hours Maximum Marks: 80 Date: 20 .01.2025 SECTION A (40 Marks) (Attempt all questions from this Section.) MARKING SCHEME Question 1 (c) Both P and Q (i) 1 (ii) (d) Iron(II) ion 1 (iii) (a) 4.48 litre 1 (iv) (b) Both positive and negative ions 1 (v) (a) Both A and R are true and R is the correct explanation of A 1 (vi) (d) 3 shells and 2 valence electrons 1 (vii) (b) Compound B 1 (viii) (c) 19 1 (ix) (b) Zinc sulphate 1 (x) (a) 2,1 1 (xi) (c) A is true but R is false 1 (xii) (b) 1:4 1 (xiii) (d) Aqueous potassium chloride 1 (xiv) (b) Nitrate 1 (xv) (c) They react with both acids and bases to form salt and water 1 Question 2 (a) Left electrode is known as oxidizing electrode because copper atom loses electrons at this electrode. i.e; it is called anode. 2 1 (b) At anode: Cu 2e- Cu2+ 2 (c) Reddish brown copper is deposited at cathode and anode mass depleted/ colour of the solution does not change. 5 (ii) (a) Sulphuric acid - contact process (b) Electroplating - Coating of a metal on cathode (c) Carbon tetrachloride - Covalent compound (d) Complete shell - Noble gases (e) Alkali metals - K, Na, Li, Rb (iii) (a) smallest (b) nitric acid (c) reddish brown (d) substitution (e) neutral (iv) (a) covalent bond (b) sulphur dioxide (c) alloy (d) electron affinity (e) catenation (v) (a) (1) CH3 CH2 CHO (2) CH3 CH2 =CH2 CH3 (3) CH3 OH (b) (1) ethanoic acid (2) 2- butanol 5 5 3 2 SECTION B (40 Marks) (Attempt any four questions) Question 3 Expected Answer Marks (i) (a) B will ionise more easily due to high electron affinity and hence has more tendency to accept electron. (b) Electron affinity decreases down the group (c) Neon 3 (ii) (a) (b) (c) (d) 4 (iii) (a)Oxygen (b)SO3 + H2SO4= H2S2O7 Oleum Question 4 (i) (ii) (iii) Neutralisation by titration Precipitation Neutralisation Displacement 200 4 (a) KNO3 + H2SO4 KHSO4 + HNO3 (b) Hard crust of potassium sulphate is formed which is difficult to remove/ nitric acid decomposes to form reddish brown fumes of N2O and water (c) Cu + 4HNO3 cone. Cu (NO3)2 2 + 2H2O + 2NO2. Reddish brown fumes of nitrogen dioxide/ solution turns greenish blue (a) Catalyst 4 (b) Reaction is exothermic (c) Water (d) NO (a) To prevent oxidation at anode by oxygen evolved/ minimizes or prevents heat loss by radiation. (b) Cathode Question 5 (i) (ii) 2 5 (a) (b) (c) (d) (e) 200 900 450 N2 + 3H2 2NH3 Haber s process 300 litres 200 litres Reaction proceeds with evolution of heat (a) No, because melting point of copper is very high 2 (b) Solder (iii) Question 6 (i) (a) Cu2+ (b) CO2 (c) CuCO3 3 3 (a) (iii) (b) Halogens (c) A (ii) (a) Nickel sulphate (b) Piece of pure nickel metal (c) Ni(s) Ni2+ +2e- 3 (iii) (a) Chloride (b) Sulphite 2 (iv) (a) Metal M 2 (b) Ions 3 Question 7 2NH3 + H2SO4 [NH4]2SO4 (i) 2(22.4 lit.) 2[14+(4 1)]+32+(4 16)=132g 132 g of [NH4]2SO4 is produced by 2 x 22.4 lit of NH 3 59 g of ammonium sulphate is produced by 2 22.4 x59 132 = 20.02 lit. of ammonia 2C2H6 (ii) + 7O2 2x22400 7x22400 cc =300 =1050 4CO2 + 6H2O (a) 2x22400 cc of ethane gives 4x22400 cc CO2 300 cc ethane will give 600 cc CO2 (b) Oxygen, For 300 cc ethane 1050 cc of O2 will be required. 3 So unused O2 left = 1250-1050= 200 cc (c) Gay Lussac s law (iii) (a) Solution becomes acidic and pH will be below 7 2 (b) Solution will be basic and pH of the solution will be above 7 (iv) Question 8 (i) 2 2 (a) Ni (b)structure of C2H6 (ii) 2 (a) C2H6 + Na2CO3 (b) C2H2 +Ca (OH)2 + C2H2 (iii) 2 (a) C2H6+ Na2CO3 (b) Y= CH3COOC2H5, z= H2SO4 (iv) 4 (a) Alkanes and alkenes (b) H2C=CH2 (c) Substitution (d) Reddish brown to colourless

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