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#molar mass

93 public questions tagged with this topic.

Mean Free Path and Molecular Diameter

At STP, 22.4 litres of oxygen gas (O₂) is present. What is the mass of this gas? (Molecular mass of O₂ = 32 u, 1 u = 1 g

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. At STP (273 K, 1 atm), 1 mole of any ideal gas occupies 22.4 litres (molar volume).Given volume = 22.4 litres, so number of moles (μ) = (22.4)/(22.4) = 1 mol .Mass = μ × molecular mass = 1 × 32 = 32 g . Substituting values gives 32 g, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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CHEMISTRY

A solution of 6.8 g of a non-volatile solute in 200 g of water has a boiling point elevation of 0.26 K. What is the mola

Given: A solution of 6.8 g of a non-volatile solute in 200 g of water has a boiling point elevation of 0.26 K. What is the molar mass of the solute? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Molality = Δ T_b/K_b = 0.26/0.52 = 0.5 mol/kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles = 0.5 × 0.2 = 0.1 mol . Molar mass = 6.8/0.1 = 68 g/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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the mass of one mole of ethanol (Câ‚‚Hâ‚…OH)? (Atomic masses: C = 12, H = 1, O = 16)

Given: What is the mass of one mole of ethanol (C₂H₅OH)? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Molar mass = (2 × 12) + (6 × 1) + 16 = 46 g/mol.. 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

The vapor pressure of pure water is 25 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor

(p⁰ - p/p⁰) = xsolute . (25 - 23/25) = 0.08 . Moles of water = (180/18) = 10 . xsolute = (nsolute/nsolute + 10) = 0.08 . nsolute = 0.08 (nsolute + 10) , nsolute - 0.08 nsolute = 0.8 , 0.92 nsolute = 0.8 , nsolute ≈ 0.8696 . Mass = 0.8696 × 60 ≈ 52.18 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

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Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

The vapor pressure of pure water is 28 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor

(p⁰ - p/p⁰) = xsolute . (28 - 26.6/28) = (1.4/28) = 0.05 . Moles of water = (360/18) = 20 . xsolute = (nsolute/nsolute + 20) = 0.05 . nsolute = 0.05 (nsolute + 20) , 0.95 nsolute = 1 , nsolute ≈ 1.0526 . Mass = 1.0526 × 50 ≈ 52.63 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

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