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#chemistry

1219 public questions tagged with this topic.

The number of valence electrons in Si and Ge atoms is:

**Energy bands in semiconductors** consist of valence band filled at 0 K and conduction band empty, gap E_g small ~1 eV (Si 1.1 eV, Ge 0.7 eV), insulators large gap >3 eV (C diamond 5.4 eV), conductors overlapping. Intrinsic semiconductor at 0 K behaves as insulator because no thermal excitation, at T>0 K electrons jump to conduction band leaving holes, conductivity increases with temperature. Si (third orbit) and Ge (fourth orbit) are group IV elements, each with four valence electrons (2s and 2p for Si, 4s and 4p for Ge), forming covalent bonds in their lattice. Substituting values gives 4,

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

A gas occupies 44.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Number of moles (μ) = VolumeMolar volume.μ = (44.8)/(22.4) = 2.0 mol. Substituting values gives 2.0 mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

What is the volume of 0.3 moles of an ideal gas at 2.5 atm and 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. PV = μ R T, V = (μ R T)/(P).T = 127 + 273 = 400 K, P = 2.5 × 1.01 × 10⁵ = 2.525 × 10⁵ Pa.V = (0.3 × 8.31 × 400)/(2.525 × 10⁵) = 3.95 × 10⁻³ m³ ≈ 3.95 litres. Substituting values gives 3.95 litres, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

What is the pressure of 0.4 moles of an ideal gas in a 4-litre container at 327°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. PV = μ R T, P = (μ R T)/(V).T = 327 + 273 = 600 K, V = 4 × 10⁻³ m³.P = (0.4 × 8.31 × 600)/(4 × 10⁻³) = 4.986 × 10⁵ Pa ≈ 5.0 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 5.0 atm, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres)

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Number of moles (μ) = VolumeMolar volume.μ = (5.6)/(22.4) = 0.25 mol. Substituting values gives 0.25 mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

A gas occupies 56.0 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. Number of moles (μ) = VolumeMolar volume.μ = (56.0)/(22.4) = 2.5 mol. Substituting values gives 2.5 mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

A gas occupies a volume of 11.2 litres at STP. How many molecules are present in this gas? (N_A = 6.02 × 10²³ mol⁻¹)

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Molar volume at STP = 22.4 litres/mol.Number of moles (μ) = (11.2)/(22.4) = 0.5 mol .Number of molecules = μ × N_A = 0.5 × 6.02 × 10²³ = 3.01 × 10²³ . Substituting values gives 3.01 × 10²³, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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

What is the total internal energy of 1 mole of a triatomic gas at 300 K with no vibrational modes? (R = 8.31 J mol⁻¹ K⁻¹

**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. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).U = 3 R T = 3 × 8.31 × 300 = 7479 J ≈ 7.48 kJ. Substituting values gives 7.48 kJ, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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

A solid has a molar specific heat capacity of 24.4 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. C = f × (R)/(2), 24.4 = f × (8.31)/(2).f = (24.4 × 2)/(8.31) ≈ 5.87 ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R

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

What is the number of unpaired electrons in [Cr(Hâ‚‚O)6]^{3+ ?

Given: What is the number of unpaired electrons in [Cr(Hâ‚‚O)6]^{3+ ? These values define the system as per NCERT data. Formula: Cr³⁺ ( d³ ) with Hâ‚‚O in an octahedral field has 3 unpaired electrons ( t_{2g³ ), as it’s a d³ system.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CHâ‚‚NHâ‚‚ 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⁻Â

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.

How many optical isomers are possible for [Cr(Câ‚‚Oâ‚„)3]^{3- ?

[Cr(Câ‚‚Oâ‚„)3]^{3- is octahedral with three bidentate oxalate ligands, lacking symmetry, and forms two optical isomers (d and l enantiomers). 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.