CHEMISTRY

CHEMISTRY

What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ?

Given: What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt2(2+2) = sqrt8 approx 2.83 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ni²⁺ ( d⁸ ) with NH₃ (moderate field) in an octahedral field is high spin ( t_{2g⁶ e_g² ), with 2 unpaired electrons. 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.

View Details

CHEMISTRY

Calculate the mole fraction of methanol ( CH₃OH ) in a solution containing 32 g of methanol and 144 g of water. (Molar

Given: Calculate the mole fraction of methanol ( CH₃OH ) in a solution containing 32 g of methanol and 144 g of water. (Molar mass: CH₃OH = 32 g/mol, H₂O = 18 g/mol ) These values define the system as per NCERT data. Formula: Moles of methanol = 32/32 = 1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of water = 144/18 = 8 mol . Total moles = 1 + 8 = 9. Mole fraction of methanol = 1/9 = 0.111 . 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.

View Details

CHEMISTRY

The K_a of chloroacetic acid is 1.35 × 10⁻³. What is the pH of a 0.05 M solution?

Given: The K_a of chloroacetic acid is 1.35 × 10⁻³. What is the pH of a 0.05 M solution? These values define the system as per NCERT data. Formula: K_a = x²/0.05, 1.35 × 10⁻³= x²/0.05, x² = 6.75 × 10⁻⁵. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: x = sqrt6.75 × 10⁻⁵ approx 8.22 × 10⁻³, pH = -log(8.22 × 10⁻³) approx 2.09 . 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.

View Details

CHEMISTRY

What is the maximum number of emission lines when an electron drops from n = 5 to the ground state in a hydrogen atom?

Given: What is the maximum number of emission lines when an electron drops from n = 5 to the ground state in a hydrogen atom? These values define the system as per NCERT data. Formula: Number of emission lines = n(n-1)/2, where n = 5. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: So, 5 × 4/2 = 10. 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.

View Details

CHEMISTRY

The rate constant of a reaction is 5.0 × 10⁻⁵ s^{-1 at 290 K and 1.5 × 10⁻⁴ s^{-1 at 300 K. What is the activa

Given: The rate constant of a reaction is 5.0 × 10⁻⁵ s^{-1 at 290 K and 1.5 × 10⁻⁴ s^{-1 at 300 K. What is the activation energy in kJ mol^{-1 ? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: log k_2/k_1 = E_a/2.303R ( T_2 - T_1/T_1 T_2 ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: log frac1.5 × 10⁻⁴⁵.0 × 10⁻⁵= log 3 = 0.477 . 0.477 = E_a/2.303 × 8.314 ( 10/290 × 300 ), E_a approx 51.5 kJ mol^{-1 . 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.

View Details

CHEMISTRY

Which of the following amines forms a salt soluble in excess NaOH when treated with benzenesulphonyl chloride?

Primary amines like CH₃NH₂ form sulphonamides with benzenesulphonyl chloride that have an acidic hydrogen, making them soluble in excess NaOH in the Hinsberg test.

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

View Details

CHEMISTRY

What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity i

Given: What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity is 0.14 × 10⁻² S cm⁻¹? These values define the system as per NCERT data. Formula: Cell constant, G^* = kappa × R = 0.14 × 10⁻² × 200 = 0.28 cm^{-1 .. 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: Solutions, Topic: Concentration terms and colligative properties.

View Details

CHEMISTRY

What is the molality of a solution containing 5 g of urea (NHâ‚‚CONHâ‚‚) in 250 g of water? (Molar mass of urea = 60 g/m

Given: What is the molality of a solution containing 5 g of urea (NH₂CONH₂) in 250 g of water? (Molar mass of urea = 60 g/mol) These values define the system as per NCERT data. Formula: Moles = 5 / 60 = 0.0833 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of solvent = 0.25 kg. Molality = 0.0833 / 0.25 ≈ 0.333 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.

View Details

CHEMISTRY

Calculate the osmotic pressure of a solution containing 2.7 g of glucose ( C₆H₁₂O₆ ) in 250 mL of water at 27°C

Given: Calculate the osmotic pressure of a solution containing 2.7 g of glucose ( C₆H₁₂O₆ ) in 250 mL of water at 27°C. ( R = 0.0821 L atm/mol K, Molar mass of glucose = 180 g/mol ) These values define the system as per NCERT data. Formula: Moles of glucose = 2.7/180 = 0.015 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Volume = 0.25 L. Molarity = 0.015/0.25 = 0.06 M . Pi = 0.06 × 0.0821 × 300 = 1.478 atm . 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.

View Details

CHEMISTRY

What is the molar conductivity at infinite dilution for NaCl if the ionic molar conductivities of Na⁺ and Cl⁻ are 50

Given: What is the molar conductivity at infinite dilution for NaCl if the ionic molar conductivities of Na⁺ and Cl⁻ are 50.1 S cm² mol⁻¹ and 76.3 S cm² mol⁻¹ respectively? These values define the system as per NCERT data. Formula: Kohlrausch’s law: Lambda_m⁰ = lambda_{Na^{+⁰ + lambda_{Cl^{-⁰. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Lambda_m⁰ = 50.1 + 76.3 = 126.4 S cm² mol^{-1 . 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.

View Details

CHEMISTRY

How many grams of N₂ are required to produce 34 g of NH₃ in the reaction N₂ + 3H₂ -> 2NH₃ ? (Atomic masses: N

Given: How many grams of N₂ are required to produce 34 g of NH₃ in the reaction N₂ + 3H₂ -> 2NH₃ ? (Atomic masses: N = 14, H = 1) These values define the system as per NCERT data. Formula: Molar mass of NH₃ = 17 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of NH₃ = 34 / 17 = 2 mol. 1 mol N₂ produces 2 mol NH₃. Moles of N₂ = 1 mol. Mass = 1 × 28 = 28 g. 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.

View Details

CHEMISTRY

For the reaction H₂(g) + Cl₂(g) -> 2HCl(g), if Δ H = -184.6 kJ/mol at 298 K, what is Δ U (R = 8.314 J/mol · K)?

Given: For the reaction H₂(g) + Cl₂(g) -> 2HCl(g), if Δ H = -184.6 kJ/mol at 298 K, what is Δ U (R = 8.314 J/mol · K)? These values define the system as per NCERT data. Formula: Δ n_g = 2 - (1 + 1) = 0, so Δ H = Δ U + Δ n_g RT = Δ U. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Δ U = -184.6 kJ/mol. 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.

View Details