This category collects chemistry practice questions spanning physical, organic and inorganic topics. Work through them one at a time to check your understanding of reactions, formulas and core concepts, or use them as a quick revision round before a test. Each attempt shows you which areas are solid and which need another look.
[PtCl₂(NH₃)2] is square planar and can form cis and trans isomers, unlike tetrahedral or fully symmetric octahedral complexes. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: A zero-order reaction has a half-life of 25 minutes with an initial concentration of 0.25 M. What is the rate constant in mol L^{-1 min^{-1 ? These values define the system as per NCERT data. Formula: t_{1/2 = frac[R]_02k, 25 = 0.25/2k, k = 0.25/50 = 0.005 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: A reaction follows the rate law Rate = k[A][B] . If k = 0.5 L mol^{-1 s^{-1, [A] = 0.2 M, and [B] = 0.3 M, what is the initial rate in mol L^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k[A][B] = 0.5 × 0.2 × 0.3 = 0.03 mol L^{-1 s^{-1 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: For the reaction N₂(g) + O₂(g) 2NO(g), K_c = 0.1 at 2000 K. If 0.5 mol of N₂ and 0.5 mol of O₂ are placed in a 5 L vessel, what is the equilibrium concentration of NO ? These values define the system as per NCERT data. Formula: Initial [N₂] = [O₂] = 0.5 / 5 = 0.1 M. This is standard NCERT relation. Substitution & Calculation: Let [NO] = 2x at equilibrium. Then, [N₂] = [O₂] = 0.1 - x . K_c = frac[NO]²[N₂][O₂] = (2x)²/(0.1 - x)² = 0.1 . 4x²/(0.1 - x)² = 0.1, 2x/0.1 - x = sqrt0.1 approx 0.316 . Solving: 2x = 0.0316 - 0.316x, 2.316x = 0.0316, x approx 0.0136, [NO] = 2x approx
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: Calculate the boiling point elevation of a solution containing 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 450 g of water. ( K_b = 0.52 K kg/mol, Molar mass of sucrose = 342 g/mol ) These values define the system as per NCERT data. Formula: Moles of sucrose = 17.1/342 = 0.05 mol. This is standard NCERT relation. Substitution & Calculation: Molality = 0.05/0.45 = 0.111 mol/kg . Δ T_b = 0.52 × 0.111 = 0.058 K . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Boiling point elevation ΔT_b = K_b·m, urea NH₂CONH₂ example, colligative properties.
Given: What is the molality of a solution containing 10 g of sugar (molar mass = 342 g/mol) dissolved in 500 g of water? These values define the system as per NCERT data. Formula: Moles = 10 / 342 ≈ 0.0292 mol. This is standard NCERT relation. Substitution & Calculation: Molality = 0.0292 / 0.5 ≈ 0.0584 m. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: During the electrolysis of aqueous CuSO₄ using copper electrodes, what mass of copper is deposited at the cathode if a current of 1.5 A flows for 10 minutes? (Molar mass of Cu = 63 g/mol, F = 96500 C/mol) These values define the system as per NCERT data. Formula: Charge, Q = I × t = 1.5 × 600 = 900 C. This is standard NCERT relation. Substitution & Calculation: For Cu²⁺ + 2e⁻ → Cu(s), 2F (2 × 96500 C) deposits 63 g of Cu. Mass = 63 × 900/2 × 96500 approx 0.294 g . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Electrochemistry, Topic: Faraday's first law, charge to reduce Al³⁺ to Al, 3F = 3 × 96500 C.
Given: For C(s) + H₂O(g) CO(g) + H₂(g), K_p = 0.4 atm at 1000 K. If initial p_{H₂O = 1 atm, what is p_{CO at equilibrium? These values define the system as per NCERT data. Formula: Let p_{CO = p_{H₂ = x, p_{H₂O = 1 - x. This is standard NCERT relation. Substitution & Calculation: K_p = fracp_{CO · p_{H₂p_{H₂O = x²/1 - x = 0.4 . x² = 0.4 - 0.4x, x² + 0.4x - 0.4 = 0, x approx 0.44 atm. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the osmotic pressure of a solution containing 1.11 g of CaCl₂ in 300 mL of water at 27°C, assuming complete dissociation? ( R = 0.0821 L atm/mol K, Molar mass of CaCl₂ = 111 g/mol, i = 3 ) These values define the system as per NCERT data. Formula: Moles of CaCl₂ = 1.11/111 = 0.01 mol. This is standard NCERT relation. Substitution & Calculation: Molarity = 0.01/0.3 = 0.0333 M . Pi = i · C · R · T = 3 × 0.0333 × 0.0821 × 300 = 2.46 atm . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The K_{sp of FeCO₃ is 3.1 × 10⁻¹¹. What is its solubility in water? These values define the system as per NCERT data. Formula: FeCO₃(s) Fe²+ + CO₃²-, K_{sp = S². This is standard NCERT relation. Substitution & Calculation: S² = 3.1 × 10⁻¹¹, S = sqrt3.1 × 10⁻¹¹approx 5.57 × 10⁻⁶M. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
The carbon in the C#N group is sp hybridized, forming a linear shape with a bond angle of 180°. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
In CH₂=CH₂ (ethene), each carbon is sp² hybridized, forming a trigonal planar shape with bond angles of 120°. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
