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CHEMISTRY

A set of chemistry practice questions spanning concepts, reactions and calculations across physical, organic and inorganic chemistry. Attempt them topic by topic or in one sitting, and use the results to decide where your revision time is best spent.

45 questions

CHEMISTRY

A gaseous reaction follows Rate = k p_A² . If k = 0.1 atm^{-1 s^{-1 and p_A = 0.4 atm, what is the rate in atm s^{-1 ?

Given: A gaseous reaction follows Rate = k p_A² . If k = 0.1 atm^{-1 s^{-1 and p_A = 0.4 atm, what is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A² = 0.1 × (0.4)² = 0.1 × 0.16 = 0.016 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

The initial rate of a reaction is 2.0 × 10⁻³ mol L^{-1 s^{-1 when [A] = 0.5 M and [B] = 0.2 M. If the rate law is Ra

Given: The initial rate of a reaction is 2.0 × 10⁻³ mol L^{-1 s^{-1 when [A] = 0.5 M and [B] = 0.2 M. If the rate law is Rate = k[A][B], what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A][B] = frac2.0 × 10⁻³⁰.5 × 0.2 = 0.02 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

How many electrons are in the L shell of an atom with 10 electrons?

Given: How many electrons are in the L shell of an atom with 10 electrons? These values define the system as per NCERT data. Formula: The L shell (n = 2) has 2s² 2p⁶ = 2 + 6 = 8 electrons.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 10 electrons (neon): 1s² 2s² 2p⁶. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

The K_a of phenol is 1.0 × 10⁻¹⁰. What is the pH of a 0.01 M solution?

Given: The K_a of phenol is 1.0 × 10⁻¹⁰. What is the pH of a 0.01 M solution? These values define the system as per NCERT data. Formula: K_a = x²/0.01, 1.0 × 10⁻¹⁰= x²/0.01, x² = 1.0 × 10⁻¹². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: x = sqrt1.0 × 10⁻¹²= 1.0 × 10⁻⁶, pH = -log(1.0 × 10⁻⁶) = 6.0 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

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CHEMISTRY

The rate law of a reaction is Rate = k[A][B] . If both [A] and [B] are increased by a factor of 3, by what factor does t

Given: The rate law of a reaction is Rate = k[A][B] . If both [A] and [B] are increased by a factor of 3, by what factor does the rate increase? These values define the system as per NCERT data. Formula: Initial rate = k[A][B], new rate = k(3[A])(3[B]) = 9k[A][B]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Factor = 9. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

What is the pH of a 0.006 M KOH solution, assuming complete dissociation?

Given: What is the pH of a 0.006 M KOH solution, assuming complete dissociation? These values define the system as per NCERT data. Formula: [OH-] = 0.006 M, pOH = -log(0.006) approx 2.22. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: pH = 14 - 2.22 = 11.78 approx 11.8 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

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CHEMISTRY

Which complex is diamagnetic?

[Ni(CN)4]^{2- (Ni²⁺, d⁸ ) with strong field CN^- in a square planar geometry is low spin, with 0 unpaired electrons, making it diamagnetic. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

What is the freezing point depression of a solution containing 3.6 g of glucose ( C₆H₁₂O₆ ) in 100 g of water? (

Given: What is the freezing point depression of a solution containing 3.6 g of glucose ( C₆H₁₂O₆ ) in 100 g of water? ( K_f = 1.86 K kg/mol, Molar mass of glucose = 180 g/mol ) These values define the system as per NCERT data. Formula: Moles of glucose = 3.6/180 = 0.02 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.02/0.1 = 0.2 mol/kg . Δ T_f = 1.86 × 0.2 = 0.372 K . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the molality of a solution containing 6 g of urea (NHâ‚‚CONHâ‚‚) in 150 g of water? (Molar mass of urea = 60 g/m

Given: What is the molality of a solution containing 6 g of urea (NH₂CONH₂) in 150 g of water? (Molar mass of urea = 60 g/mol) These values define the system as per NCERT data. Formula: Moles = 6 / 60 = 0.1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of solvent = 0.15 kg. Molality = 0.1 / 0.15 ≈ 0.667 m. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the EMF of the ll Pb(s) | Pb²⁺(0.05 M) || Ag⁺(0.002 M) | Ag(s) at 298 K, given E_{Pb^{2+/Pb⁰ = -0.13 V an

Given: What is the EMF of the ll Pb(s) | Pb²⁺(0.05 M) || Ag⁺(0.002 M) | Ag(s) at 298 K, given E_{Pb^{2+/Pb⁰ = -0.13 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.13) = 0.93 V, n = 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Pb^{2+][Ag^{+]² . Q = 0.05/(0.002)² = 12500, log Q = 4.0969 . E_{ll = 0.93 - 0.059/2 × 4.0969 = 0.93 - 0.1209 = 0.8091 V approx 0.81 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

A 1.2 M solution of KCl has a density of 1.1 g/mL. What is its molality? (Molar mass of KCl = 74.5 g/mol)

Given: A 1.2 M solution of KCl has a density of 1.1 g/mL. What is its molality? (Molar mass of KCl = 74.5 g/mol) These values define the system as per NCERT data. Formula: Mass of 1 L = 1100 g. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of KCl = 1.2 × 74.5 = 89.4 g. Mass of water = 1100 - 89.4 = 1010.6 g = 1.0106 kg. Molality = 1.2 / 1.0106 ≈ 1.187 m ≈ 1.19 m. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

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