JEE Chemistry - MCQ Practice Questions

In presence of 0.1 M Cl⁻: Ksp = [Ag⁺][Cl⁻]; 1.8 × 10⁻¹⁰ = s × 0.1; s = 1.8 × 10⁻⁹ M

H₂O has maximum boiling point due to strong hydrogen bonding compared to other hydrides in group 16

Cu²⁺ requires 2 electrons per atom. 1 Faraday = 1 mol of electrons; moles of Cu = $\frac{1}{2}$ = 0.5 mol

ΔG° = -nFE°; For bivalent ion transfer: ΔG° = -2 × 96485 × 0.92 ≈ -177.6 kJ/mol

Benzene and toluene are both aromatic hydrocarbons with similar molecular structures, forming nearly ideal solutions

For reversible adiabatic process: q = 0 (adiabatic) and dS_surr = 0, so ΔS_total = 0, therefore ΔS_system = 0

New rate = k(2[A])²(0.5[B]) = k × 4[A]² × 0.5[B] = 2k[A]²[B] = 2 × original rate

Chemical potential μᵢ = (∂G/∂nᵢ)T,P, which is the partial molar Gibbs free energy

ΔG = ΔH - TΔS; For spontaneity ΔG < 0. Exothermic (ΔH < 0) with ΔS > 0 always gives ΔG < 0

At higher dilutions, weak electrolytes ionize more completely, increasing degree of ionization and thus molar conductivity

Diamond has a 3D tetrahedral covalent network structure with C-C bonds in all directions, while graphite has layered structure with weak van der Waals forces between layers

Initial: 2 atm of A. At equilibrium with α = 0.5: A dissociates to (1-0.5)×2 = 1 atm, producing B and C each at 0.5 atm. Total = 1 + 0.5 + 0.5 = 2 atm × (1 + α) = 2 × 1.5 = 3 atm

Kc depends only on temperature. Pressure, concentration changes, and catalysts do not alter Kc value—they shift equilibrium position but not the constant itself.

For weak acid: α = √(Ka/C) = √(1×10⁻⁵/0.1) = √(1×10⁻⁴) = 0.01 or 1%. Using exact formula: α ≈ 0.0316 (3.16%)

ΔG° = -20 kJ/mol (negative, spontaneous). Since Q (0.01) < Kp (1000), reaction shifts forward. ΔG = ΔG° + RTlnQ = -20 + 8.314×298×ln(0.01) will be more negative, confirming forward reaction.

Cp - Cv = R = 8.314 J/(mol·K). Therefore, Cv = 29 - 8.314 ≈ 20.7 ≈ 20.8 J/(mol·K). This indicates a diatomic gas (Cp = $\frac{7}{2}$ R, Cv = $\frac{5}{2}$ R).

For first-order reaction, each half-life reduces concentration by 50%. After 1st half-life (10 min): 50% remains. After 2nd half-life (20 min): 25% remains = $\frac{1}{4}$th. Total time = 2 × 10 = 20 minutes.