CHEMISTRY

CHEMISTRY

What is the oxidation state of iron in K₄[Fe(CN)6] ?

Given: What is the oxidation state of iron in K₄[Fe(CN)6] ? These values define the system as per NCERT data. Formula: Each CN^- is -1, 6 ligands = -6. This is standard NCERT relation. Substitution & Calculation: The complex [Fe(CN)6]^{4- is balanced by 4 K⁺, so Fe’s oxidation state is x - 6 = -4, x = +2 . 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.

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CHEMISTRY

What is the maximum oxidation state exhibited by vanadium in the 3d series?

Given: What is the maximum oxidation state exhibited by vanadium in the 3d series? These values define the system as per NCERT data. Formula: Vanadium (V, Z = 23) exhibits oxidation states from +2 to +5, with +5 being the maximum, as seen in VO₂^+ .. 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.

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CHEMISTRY

A gaseous reaction follows Rate = k p_A p_B with k = 0.04 atm^{-1 s^{-1, p_A = 0.5 atm, and p_B = 0.2 atm . What is the

Given: A gaseous reaction follows Rate = k p_A p_B with k = 0.04 atm^{-1 s^{-1, p_A = 0.5 atm, and p_B = 0.2 atm . What is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A p_B = 0.04 × 0.5 × 0.2 = 0.004 .. 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.

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CHEMISTRY

How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5

Given: How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5 g/mol) These values define the system as per NCERT data. Formula: Moles of Cu = 0.635/63.5 = 0.01 mol. This is standard NCERT relation. Substitution & Calculation: Reaction: Cu²⁺ + 2e⁻ → Cu(s), 2F deposits 63.5 g. . Faradays = 0.01 × 2 = 0.02 F . 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.

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CHEMISTRY

The K_p for CS₂(g) + 4H₂(g) CH₄(g) + 2H₂S(g) is 0.25 at 900 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ?

Given: The K_p for CS₂(g) + 4H₂(g) CH₄(g) + 2H₂S(g) is 0.25 at 900 K. What is K_c if R = 0.0821 L · atm · mol^{-1 · K^{-1 ? These values define the system as per NCERT data. Formula: Δ n = (1 + 2) - (1 + 4) = -2, K_p = K_c (RT)^{Δ n. This is standard NCERT relation. Substitution & Calculation: RT = 0.0821 × 900 = 73.89, (RT)^{-2 = (73.89)^{-2 approx 1.83 × 10⁻⁴. 0.25 = K_c · 1.83 × 10⁻⁴, K_c approx 1365 . 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.

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CHEMISTRY

A solution of two volatile liquids A and B has a total vapor pressure of 320 mm Hg. If the vapor pressure of pure A is 4

Given: A solution of two volatile liquids A and B has a total vapor pressure of 320 mm Hg. If the vapor pressure of pure A is 400 mm Hg and that of pure B is 200 mm Hg, what is the mole fraction of B in the solution? These values define the system as per NCERT data. Formula: Using Raoult's law: p_{total = x_A p_A⁰ + x_B p_B⁰, where x_A + x_B = 1. This is standard NCERT relation. Substitution & Calculation: 320 = (1 - x_B) · 400 + x_B · 200 . 320 = 400 - 400 x_B + 200 x_B . 320 = 400 - 200 x_B, 200 x_B = 80, x_B = 0.4 . 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.

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CHEMISTRY

A solution contains 10 g of urea ( NH₂CONH₂ ) in 190 g of water. What is the molality of the solution? (Molar mass: NH₂C

Given: A solution contains 10 g of urea ( NH₂CONH₂ ) in 190 g of water. What is the molality of the solution? (Molar mass: NH₂CONH₂ = 60 g/mol ) These values define the system as per NCERT data. Formula: Moles of urea = 10/60 = 0.1667 mol. This is standard NCERT relation. Substitution & Calculation: Mass of solvent in kg = 190/1000 = 0.19 kg . Molality = 0.1667/0.19 = 0.877 mol/kg . 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.

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CHEMISTRY

How many grams of H₂O are produced from 10 g of H₂SO₄ in the reaction H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O ? (Atomic masses: H

Given: How many grams of H₂O are produced from 10 g of H₂SO₄ in the reaction H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O ? (Atomic masses: H = 1, S = 32, O = 16) These values define the system as per NCERT data. Formula: Moles of H₂SO₄ = 10 / 98 ≈ 0.102 mol. This is standard NCERT relation. Substitution & Calculation: 1 mol H₂SO₄ produces 2 mol H₂O. Moles of H₂O = 0.102 × 2 ≈ 0.204 mol. Mass = 0.204 × 18 ≈ 3.67 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: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

For a reaction with Δ H = -50 kJ/mol and Δ S = 0.25 kJ/K · mol, at what temperature is Δ G = 0?

Given: For a reaction with Δ H = -50 kJ/mol and Δ S = 0.25 kJ/K · mol, at what temperature is Δ G = 0? These values define the system as per NCERT data. Formula: At equilibrium, Δ G = 0, so Δ H - TΔ S = 0. This is standard NCERT relation. Substitution & Calculation: T = Δ H / Δ S = -50 / 0.25 = 200 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: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

What is the mass percentage of magnesium in Mg(NO₃)₂? (Atomic masses: Mg = 24, N = 14, O = 16)

Given: What is the mass percentage of magnesium in Mg(NO₃)₂? (Atomic masses: Mg = 24, N = 14, O = 16) These values define the system as per NCERT data. Formula: Molar mass = 24 + (2 × 14) + (6 × 16) = 24 + 28 + 96 = 148 g/mol. This is standard NCERT relation. Substitution & Calculation: Mass of Mg = 24 g. % Mg = (24 / 148) × 100 ≈ 16.22%. 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.

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CHEMISTRY

How many geometrical isomers are possible for [Pt(NH₃)(Br)(Cl)(py)] ?

For a square planar [Ma b c d] complex, 3 geometrical isomers are possible by varying the relative positions of the four different ligands. 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.

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CHEMISTRY

What is the coefficient of H₂O when the reaction MnO₄^- + I^- -> MnO₂ + I₂ is balanced in basic medium?

Using the half-reaction method: Oxidation: 2I^- -> I₂ + 2e^- . Reduction: MnO₄^- + 2H₂O + 3e^- -> MnO₂ + 4OH^- . Equalize electrons (6), combine: 2MnO₄^- + 6I^- + 4H₂O -> 2MnO₂ + 3I₂ + 8OH^- . Coefficient of H₂O is 4.

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.

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