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CHEMISTRY

This chemistry category focuses on the ideas that come up again and again: atomic structure and bonding, periodic trends, reaction types and mechanisms, and stoichiometric calculations. Use it to firm up theory and to practise the calculation-heavy questions that cost marks when rushed.

45 questions

CHEMISTRY

What is the formula for hexaamminecobalt(III) sulphate?

Co³⁺ with 6 NH₃ forms [Co(NH₃)6]^{3+ . Sulphate ( SO₄^{2- ) requires 3 units to balance two +3 complexes, giving [Co(NH₃)6]2(SO₄)3 . This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

What is the total number of lone pairs on the nitrogen atom in the HNO₃ molecule?

In HNO₃, nitrogen forms 3 bonds (1 double to O, 1 single to O, 1 single to OH), using all 5 valence electrons, leaving 0 lone pairs. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Chemical Bonding (Latest NCERT 2026-27), Topic: Hybridization of nitrogen in CH₃NH₂, sp³ hybridization and pyramidal geometry

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CHEMISTRY

A first-order reaction is 25% complete in 15 minutes. What is the rate constant in min^{-1 ?

Given: A first-order reaction is 25% complete in 15 minutes. What is the rate constant in min^{-1 ? Formula: 25% complete, 75% remains: frac[R]_0[R] = 100/75 = 1.333. Substitution & Calculation: k = 2.303/t log frac[R]_0[R] = 2.303/15 log 1.333 = 0.0192 min^{-1 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Chemical Kinetics (Latest NCERT 2026-27), Topic: Arrhenius equation, k = A e^(-E_a/RT), activation energy and rate constant

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CHEMISTRY

How much current (in amperes) is required to deposit 0.108 g of silver from AgNO₃ solution in 965 seconds? (Molar mass o

Given: How much current (in amperes) is required to deposit 0.108 g of silver from AgNO₃ solution in 965 seconds? (Molar mass of Ag = 108 g/mol, F = 96500 C/mol) Formula: Moles of Ag = 0.108/108 = 0.001 mol. Substitution & Calculation: Reaction: Ag⁺ + e⁻ → Ag(s), 1F deposits 108 g. . Charge = 0.001 × 96500 = 96.5 C . Current = Q/t = 96.5/965 = 0.1 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Electrochemistry (Latest NCERT 2026-27), Topic: Faraday's laws, charge required to reduce Al³⁺, 3F = 3 × 96500 C

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CHEMISTRY

The K_p for NH₃(g) 1/2 N₂(g) + 3/2 H₂(g) is 0.2 atm at 500 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ?

Given: The K_p for NH₃(g) 1/2 N₂(g) + 3/2 H₂(g) is 0.2 atm at 500 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ? Formula: Δ n = (1/2 + 3/2) - 1 = 1, K_p = K_c (RT)^{Δ n. Substitution & Calculation: RT = 0.0821 × 500 = 41.05, 0.2 = K_c · 41.05, K_c = 0.2 / 41.05 approx 0.00487 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

What is the molality of a solution made by dissolving 5 g of urea (molar mass = 60 g/mol) in 200 g of water?

Given: What is the molality of a solution made by dissolving 5 g of urea (molar mass = 60 g/mol) in 200 g of water? Formula: Moles = 5 / 60 = 0.0833 mol. Substitution & Calculation: Mass of solvent = 0.2 kg. Molality = 0.0833 / 0.2 = 0.4165 ≈ 0.417 m. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII), Chapter: Solutions (Latest NCERT 2026-27), Topic: Colligative properties, ΔT_b = K_b·m, urea NH₂CONH₂ example

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CHEMISTRY

Calculate the heat required to raise the temperature of 28 g of ethanol from 30°C to 50°C (specific heat of ethanol = 2.

Given: Calculate the heat required to raise the temperature of 28 g of ethanol from 30°C to 50°C (specific heat of ethanol = 2.46 J/g · K). Formula: q = m × c × Δ T. Substitution & Calculation: m = 28 g, c = 2.46 J/g · K, Δ T = 50 - 30 = 20 K. q = 28 × 2.46 × 20 = 1377.6 J = 1.38 kJ. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is

Given: A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is the van't Hoff factor if it dimerizes? ( K_f = 5.12 K kg/mol, Molar mass = 122 g/mol ) Formula: Molality = 1.8 / 122/0.03 = 0.492 mol/kg. Substitution & Calculation: Normal Δ T_f = 5.12 × 0.492 = 2.52 K . Observed Δ T_f = 1.2 K . i = 1.2/2.52 = 0.476 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

What is the total number of pi ( π ) bonds in CH₃C#CCH₃ ?

In CH₃C#CCH₃, the triple bond (C≡C) consists of 1 sigma and 2 pi bonds. There are no other multiple bonds. Total π bonds = 2. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

What is the dissociation constant of 0.002 M methanoic acid if its molar conductivity is 40.42 S cm² mol⁻¹ and Lambda_m⁰

Given: What is the dissociation constant of 0.002 M methanoic acid if its molar conductivity is 40.42 S cm² mol⁻¹ and Lambda_m⁰ = 404.2 S cm² mol^{-1 ? Formula: α = 40.42/404.2 = 0.1. Substitution & Calculation: K_a = c α²/1 - α = 0.002 × (0.1)²/0.9 = 0.002 × 0.01/0.9 = 2.22 × 10⁻⁵mol L^{-1 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

The solubility product of Bi₂S₃ is 1.0 × 10⁻⁹⁷. What is its solubility in water?

Given: The solubility product of Bi₂S₃ is 1.0 × 10⁻⁹⁷. What is its solubility in water? Formula: Bi₂S₃(s) 2Bi³+ + 3S²-, K_{sp = (2S)² · (3S)³ = 108S⁵. Substitution & Calculation: 108S⁵ = 1.0 × 10⁻⁹⁷, S⁵ = 9.26 × 10⁻¹⁰⁰, S approx 1.32 × 10⁻²⁰M. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

The freezing point depression of a 0.1 m aqueous NaCl solution is 0.347 K. What is the van't Hoff factor? ( K_f = 1.86 K

Given: The freezing point depression of a 0.1 m aqueous NaCl solution is 0.347 K. What is the van't Hoff factor? ( K_f = 1.86 K kg/mol ) Formula: Δ T_f = i · K_f · m. Substitution & Calculation: 0.347 = i × 1.86 × 0.1 . i = 0.347/1.86 × 0.1 = 0.347/0.186 = 1.87 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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