#mass calculation

Potential Energy of Mass Systems

Two masses 2kg and 8kg are separated by 4m. What is the gravitational potential at the midpoint? (G\=6.67×10−11N m2/kg2)

Distance to midpoint: r = 2m. U = −Gm1r−Gm2r. U = −6.67×10−11(22+82). U = −6.67×10−11×(1+4) = −3.335×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -3.3 × 10⁻¹⁰ J/kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Corrosion and Applications of Electrochemistry

A mercury cell produces 0.5 g of Hg (atomic mass 200 g/mol) at the cathode. What mass of Zn (atomic mass 68 g/mol) is ox

Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ , Moles = (0.5/200) = 0.0025 mol , Charge = 0.0025 × 2 × 96500 = 482.5 C . Anode: Zn → Zn²⁺ + 2e⁻ , Moles Zn = 0.0025 mol , Mass = 0.0025 × 68 = 0.17 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the mass of urea (molar mass = 60 g/mol) required to prepare 2 kg of a 0.5 molal aqueous solution?

Molality = (Moles of solute/Mass of solvent in kg) . Moles = 0.5 × 2 = 1 mol . Mass = 1 × 60 = 60 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the mass of water required to prepare 300 g of a 5% by mass solution of a solute?

Mass of solute = (5/100) × 300 = 15 g . Mass of water = 300 - 15 = 285 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point

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Dalton's Atomic Theory and Atomic and Molecular Masses

A mixture of 12 g C and 40 g O₂ is ignited to form CO₂ and CO. If 22 g of CO₂ is produced, what is the mass of CO formed

Moles: C = 1 mol, O₂ = 1.25 mol. Moles of CO₂ = 22/44 = 0.5 mol; C used = 0.5 mol. Remaining C = 0.5 mol forms 0.5 mol CO; mass = 0.5 × 28 = 14 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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Dalton's Atomic Theory and Atomic and Molecular Masses

A mixture of 6 g H₂ and 32 g O₂ is ignited to form H₂O. What is the mass of H₂O produced? (Atomic masses: H = 1, O = 16)

Reaction: 2H₂ + O₂ → 2H₂O. Moles: H₂ = 6/2 = 3 mol, O₂ = 32/32 = 1 mol. 2 mol H₂ need 1 mol O₂; 1 mol O₂ limits, reacts with 2 mol H₂. 2 mol H₂O formed = 2 × 18 = 36 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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Dalton's Atomic Theory and Atomic and Molecular Masses

A mixture of 6 g H₂ and 44 g O₂ is ignited to form H₂O. What is the mass of H₂O produced? (Atomic masses: H = 1, O = 16)

Reaction: 2 H₂ + O₂ → 2 H₂O. Moles: H₂ = 6/2 = 3 mol, O₂ = 44/32 = 1.375 mol. 2 mol H₂ need 1 mol O₂; 1.375 mol O₂ limits, reacts with 2.75 mol H₂. 2.75 mol H₂O formed = 2.75 × 18 = 49.5 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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