CHEMISTRY

CHEMISTRY

Calculate the mole fraction of ethanol ( Câ‚‚Hâ‚…OH ) in a solution containing 69 g of ethanol and 198 g of water. (Mola

Given: Calculate the mole fraction of ethanol ( C₂H₅OH ) in a solution containing 69 g of ethanol and 198 g of water. (Molar mass: C₂H₅OH = 46 g/mol, H₂O = 18 g/mol ) These values define the system as per NCERT data. Formula: Moles of ethanol = 69/46 = 1.5 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of water = 198/18 = 11 mol . Total moles = 1.5 + 11 = 12.5. Mole fraction of ethanol = 1.5/12.5 = 0.12 . 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

In the Carius method, the percentage of sulphur in a 0.3 g compound yielding 0.699 g BaSOâ‚„ is (Atomic masses: Ba = 137

Given: In the Carius method, the percentage of sulphur in a 0.3 g compound yielding 0.699 g BaSO₄ is (Atomic masses: Ba = 137, S = 32, O = 16) These values define the system as per NCERT data. Formula: Molar mass of BaSO₄ = 137 + 32 + 64 = 233 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of S = 32 × 0.699/233 = 0.096 g. Percentage = 0.096 × 100/0.3 = 32% . 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 magnetic moment (in BM) of [CoClâ‚„]^{2- ?

Given: What is the magnetic moment (in BM) of [CoCl₄]^{2- ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt3(3+2) = sqrt15 approx 3.87 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Co²⁺ ( d⁷ ) in tetrahedral [CoCl₄]^{2- with weak field Cl^- is high spin ( e⁴ t_2³ ), with 3 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.

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CHEMISTRY

What is the bond order of O₂²- according to molecular orbital theory?

For O₂²- : 18 electrons, (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (sigma 2p_z)² (π 2p_x)² (π 2p_y)² (π^* 2p_x)² (π^* 2p_y)² . Bonding = 10, antibonding = 8. Bond order = 1/2 (10 - 8) = 1 .

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

How many geometrical isomers are possible for [Cr(NH₃)2Cl₄]^{- ?

For octahedral [Ma₂b4], 2 geometrical isomers exist: cis (two NH₃ adjacent) and trans (opposite). 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the minimum number of electrons required to fill the M shell completely?

The M shell (n = 3) has a maximum capacity of 2n² = 2 × 3² = 18 electrons, assuming all subshells (3s, 3p, 3d) are filled. 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

The rate law for a reaction is Rate = k[A]² . If the concentration of A is tripled, by what factor does the rate increa

Initial rate = k[A]², new rate = k(3[A])² = 9k[A]² . Factor = frac9k[A]²k[A]² = 9 . 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

Calculate the boiling point elevation for a solution of 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 200 g of water. (

Given: Calculate the boiling point elevation for a solution of 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 200 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 the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.05/0.2 = 0.25 mol/kg . Δ T_b = 0.52 × 0.25 = 0.13 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 EMF of a ll with Zn(s) | Zn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Zn^{2+/Z

Given: What is the EMF of a ll with Zn(s) | Zn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{H^{+/H_2⁰ = 0 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0 - (-0.76) = 0.76 V, n = 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Zn^{2+][H^{+]² . Q = 0.05/(0.1)² = 5, log Q = 0.699 . E_{ll = 0.76 - 0.059/2 × 0.699 = 0.76 - 0.0206 = 0.7394 V approx 0.74 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

Calculate the kinetic energy of an electron ejected from a metal with nu_0 = 5.0 × __10POW₁₄__Hz by light of nu = 7

Given: Calculate the kinetic energy of an electron ejected from a metal with nu_0 = 5.0 × __10POW₁₄__Hz by light of nu = 7.0 × __10POW₁₄__Hz. (h = 6.626 × 10⁻³⁴ J s) These values define the system as per NCERT data. Formula: Kinetic energy = h(nu - nu_0) = 6.626 × 10⁻³⁴ × (7.0 × __10POW₁₄__- 5.0 × __10POW₁₄__) = 6.626 × 10⁻³⁴ × 2.0 × __10POW₁₄__= 1.325 × 10⁻¹⁹ J.. 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

What is the conductivity of a 0.02 M NaCl solution at 298 K if its molar conductivity is 115 S cm² mol⁻¹?

Given: What is the conductivity of a 0.02 M NaCl solution at 298 K if its molar conductivity is 115 S cm² mol⁻¹? These values define the system as per NCERT data. Formula: Lambda_m = kappa/c × 1000, kappa = Lambda_m × c/1000. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: kappa = 115 × 0.02/1000 = 0.0023 S cm^{-1 = 2.3 × 10⁻³ S cm^{-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: Solutions, Topic: Concentration terms and colligative properties.

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CHEMISTRY

The rate constant of a reaction is 1.0 × 10⁻² s^{-1 at 27°C. If the activation energy is 60 kJ/mol, what is the rat

Given: The rate constant of a reaction is 1.0 × 10⁻² s^{-1 at 27°C. If the activation energy is 60 kJ/mol, what is the rate constant at 37°C? (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 frack_21.0 × 10⁻²= 60000/2.303 × 8.314 ( 10/300 × 310 ) = 0.336 . k_2 = 1.0 × 10⁻² × __10POW₀__.336 = 2.17 × 10⁻² s^{-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.

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