Class 12 Chemistry – Daily Revision MCQ (English Medium)
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How many particles effectively belong to one body-centred cubic (bcc) unit cell?
Chapter 1: Solid State
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Answer: C. 2
8 corner particles each contribute 1/8 (total 1) plus 1 whole particle at the body centre = 2.For a solution of a non-volatile solute, the relative lowering of vapour pressure of the solvent is equal to:
Chapter 2: Solutions
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Answer: A. the mole fraction of the solute
ΔP = P₁⁰x₂, so ΔP/P₁⁰ = x₂, the mole fraction of the solute.An aqueous solution has pH = 3 at 298 K. What is its pOH?
Chapter 3: Ionic Equilibria
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Answer: D. 11
At 298 K, pH + pOH = 14, so pOH = 14 − 3 = 11.The charge carried by one mole of electrons is called:
Chapter 5: Electrochemistry
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Answer: B. one faraday (about 96500 C)
One mole of electrons carries 96500 C; this quantity of charge is one faraday (1 F).A first order reaction has rate constant k = 0.0231 min⁻¹. Its half-life is:
Chapter 6: Chemical Kinetics
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Answer: A. 30 min
For a first order reaction t½ = 0.693/k = 0.693/0.0231 = 30 min.Ethylenediaminetetraacetate ion (EDTA⁴⁻) binds a metal ion through how many donor atoms?
Chapter 9: Coordination Compounds
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Answer: C. 6 (it is hexadentate)
EDTA⁴⁻ uses electron pairs of four oxygen and two nitrogen atoms, so it is a hexadentate ligand.An SN2 reaction taking place at a chiral carbon atom of an alkyl halide gives a product with:
Chapter 10: Halogen Derivatives
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Answer: D. inversion of configuration
In SN2 the nucleophile attacks from the side opposite to the leaving group, so the configuration flips like an umbrella (Walden inversion).Propanamide (CH₃CH₂CONH₂) is warmed with bromine and concentrated aqueous KOH. The amine formed is:
Chapter 13: Amines
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Answer: B. ethylamine (CH₃CH₂NH₂)
Hofmann bromamide degradation removes the C=O carbon, giving a primary amine with one carbon less: 3 C → 2 C, ethylamine.Which one of the following is a non-reducing sugar?
Chapter 14: Biomolecules
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Answer: C. Sucrose
In sucrose the reducing groups of both glucose (C-1) and fructose (C-2) are used in the glycosidic bond, so it is non-reducing.Conventionally, the nanometer scale is defined as the size range of:
Chapter 16: Green Chemistry and Nanochemistry
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Answer: A. 1–100 nm
The textbook defines the nanoscale as 1–100 nm, where 1 nm = 10⁻⁹ m (one billionth of a metre).
Class 12: MCQ in other subjects
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All questions are written by PadhaiNotes in our own words from the chapters of the official Maharashtra State Board Balbharati textbook (ebalbharati.in). Each question shows its chapter.