12th Standard Chemistry March 2025: Online Practice Test

Explanation. 2H₂O₂ → 2H₂O + O₂. 48 g O₂ = 48/32 = 1.5 mol per minute. Two moles of water form for every mole of O₂, so the rate of formation of water = 2 × 1.5 = 3.0 mol min⁻¹.

Explanation. K₂Cr₂O₇ + 7H₂SO₄ + 6KI → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O. One Cr₂O₇²⁻ gains 6 electrons, which oxidises 6 I⁻ to 3 I₂, so 3 moles of I₂ are liberated.

Explanation. Non-stick coating is Teflon (PTFE), made by polymerising tetrafluoroethylene, CF₂=CF₂. The paper prints "1,1,2,2-tetrafluoroethane" (as in the textbook); the intended monomer is tetrafluoroethene, so option (c) is the answer.

Explanation. Secondary amines react with nitrous acid to form N-nitrosoamines, which are yellow oily liquids. N-methylaniline is a secondary amine: C₆H₅NHCH₃ + HNO₂ → C₆H₅N(CH₃)–N=O + H₂O.

Explanation. Boric acid is a weak monobasic Lewis acid. It does not donate its own proton; the electron-deficient boron accepts OH⁻ from water: B(OH)₃ + H₂O → [B(OH)₄]⁻ + H⁺.

Explanation. Particles moving to the cathode are positively charged, so the anions cause coagulation. By the Hardy–Schulze rule, higher anion charge means greater coagulating power: [Fe(CN)₆]⁴⁻ > PO₄³⁻ > SO₄²⁻ > Cl⁻, i.e. (iii) > (ii) > (i) > (iv).

Explanation. The F–F bond is weaker than the Cl–Cl bond (about 158 vs 242 kJ mol⁻¹), so the assertion is false. The reason is also false: the small fluorine atom has greater repulsion between its lone pairs, which is why its bond is weak.

Explanation. In fluorite (CaF₂), Ca²⁺ ions form an fcc lattice and F⁻ ions occupy all tetrahedral voids. Each Ca²⁺ is surrounded by 8 F⁻ and each F⁻ by 4 Ca²⁺, giving 8 and 4.

Explanation. The sequence of α-amino acids is the primary structure. The secondary structure is the regular, hydrogen-bonded folding (fixed configuration) of the polypeptide backbone; the α-helix is only one example of it (β-sheet is another), so (b) is the complete answer.

Explanation. Ionisation of water is endothermic, so Kw increases as temperature rises. Only option (c), 2.71 × 10⁻¹⁴, is larger than 1.00 × 10⁻¹⁴.

Explanation. Br⁻ outside the bracket makes the complex ion +1. NO₂⁻ is −1, H₂O and NH₃ are neutral: x + (−1) = +1, so x = +2.

Explanation. 1 F = 96500 C is the charge of 1 mole (6.022 × 10²³) of electrons. 9650 C = 0.1 F, so number of electrons = 0.1 × 6.022 × 10²³ = 6.022 × 10²².

Explanation. The –NO₂ group at the para position stabilises the phenoxide ion by both −I and −R effects, so 4-nitrophenol is the most acidic. In 2-nitrophenol intramolecular hydrogen bonding makes the proton harder to lose, 3-nitrophenol has only the −I effect, and Cl is a much weaker electron-withdrawing group.

Explanation. CH₃Br + KCN → CH₃CN (A); acid hydrolysis gives CH₃COOH (B); PCl₅ converts the –COOH group to –COCl, giving CH₃COCl, acetyl chloride (C).

Explanation. Leaching gives the soluble complex Na[Ag(CN)₂]. Zinc, being more electropositive, displaces silver: 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag.