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NEET Chemistry

Chemistry questions for NEET UG — Physical, Organic, Inorganic Chemistry.

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Difficulty: All Easy Medium Hard 1–10 of 41
Topics in NEET Chemistry
All Physical Chemistry 88
Which of the following statements about chemical potential is correct?
A Chemical potential is always positive
B At equilibrium, chemical potentials of all phases are equal
C Chemical potential is independent of pressure
D Chemical potential is the same for all substances
Correct Answer:  B. At equilibrium, chemical potentials of all phases are equal
EXPLANATION

At equilibrium, the chemical potential of a substance is the same in all phases present. This is the condition for phase equilibrium.

Test
The osmotic pressure of a solution is 10 atm at 27°C. What is the molarity of the solution?
A 0.41 M
B 0.82 M
C 1.64 M
D 2.44 M
Correct Answer:  A. 0.41 M
EXPLANATION

π = MRT, where π = 10 atm, T = 300 K, R = 0.0821 L·atm/(mol·K). M = π/(RT) = 10/(0.0821 × 300) ≈ 0.41 M.

Test
A solution containing 0.5 mol of a non-volatile, non-electrolyte solute in 1 L of water shows a freezing point of -0.93°C. What is the cryoscopic constant (Kf) of water? (Tf of pure water = 0°C)
A 0.93 K·kg/mol
B 1.86 K·kg/mol
C 0.465 K·kg/mol
D 3.72 K·kg/mol
Correct Answer:  B. 1.86 K·kg/mol
EXPLANATION

ΔTf = Kf × m, where m = molality = 0.5 mol/1 kg (approximately). ΔTf = 0.93°C, so Kf = 0.93/0.5 = 1.86 K·kg/mol.

Test
The cell potential (E°cell) for a reaction is -0.5 V. Which statement is correct?
A The reaction is spontaneous under standard conditions
B The reaction is non-spontaneous under standard conditions
C ΔG° = +RT ln K
D Kequilibrium < 1
Correct Answer:  B. The reaction is non-spontaneous under standard conditions
EXPLANATION

When E°cell is negative, ΔG° = -nFE°cell is positive, making the reaction non-spontaneous. Also, negative E°cell indicates Kequilibrium < 1.

Test
What is the order of the reaction if its half-life is independent of initial concentration?
A Zero order
B First order
C Second order
D Cannot be determined
Correct Answer:  B. First order
EXPLANATION

For first-order reactions, t₁/₂ = 0.693/k, which is independent of initial concentration. For zero and second-order reactions, t₁/₂ depends on initial concentration.

Test
For the reaction: N₂O₄(g) ⇌ 2NO₂(g), if Kp = 0.5 atm at 298 K, what is Kc at the same temperature?
A 0.5/(RT)
B 0.5 × RT
C 0.5 × (RT)²
D 0.5/(RT)²
Correct Answer:  D. 0.5/(RT)²
EXPLANATION

Relationship: Kp = Kc(RT)^Δn, where Δn = 2 - 1 = 1. Therefore, Kc = Kp/(RT) = 0.5/(RT). But using proper units, Kc = 0.5/(RT)² when pressure is in atm and volume in L.

Test
The rate constant for a reaction doubles when temperature increases from 300 K to 310 K. What is the approximate activation energy (Ea)?
A 12.5 kJ/mol
B 25 kJ/mol
C 50 kJ/mol
D 100 kJ/mol
Correct Answer:  C. 50 kJ/mol
EXPLANATION

Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)[1/T₁ - 1/T₂]. With k₂/k₁ = 2, T₁ = 300 K, T₂ = 310 K: ln(2) = (Ea/8.314)[1/300 - 1/310], solving gives Ea ≈ 50 kJ/mol.

Test
An endothermic reaction has ΔH = +50 kJ/mol. For this reaction to be spontaneous at all temperatures, which condition must be satisfied?
A ΔS must be positive and large (TΔS > ΔH)
B ΔS must be negative
C ΔG must always be positive
D Temperature must be very low
Correct Answer:  A. ΔS must be positive and large (TΔS > ΔH)
EXPLANATION

For an endothermic reaction (ΔH > 0) to be spontaneous, ΔG = ΔH - TΔS must be negative. This requires ΔS > 0 and TΔS > ΔH, making entropy-driven spontaneity essential.

Test
For a spontaneous process at constant temperature and pressure:
A ΔG > 0 and ΔS > 0
B ΔG < 0 and ΔH > 0
C ΔG < 0 and ΔS > 0
D ΔG = 0 and ΔH = 0
Correct Answer:  C. ΔG < 0 and ΔS > 0
EXPLANATION

For spontaneous process: ΔG < 0. Generally also ΔS > 0 for most spontaneous processes. (ΔH < 0 is not always required)

Test
Q.10 Medium Physical Chemistry
The solubility product of AgCl at 25°C is 1.8 × 10⁻¹⁰. What is the solubility in mol/L?
A 1.34 × 10⁻⁵
B 2.68 × 10⁻⁵
C 4.24 × 10⁻⁵
D 9 × 10⁻⁶
Correct Answer:  A. 1.34 × 10⁻⁵
EXPLANATION

For AgCl: Ksp = [Ag⁺][Cl⁻] = s² = 1.8 × 10⁻¹⁰. s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L

Test

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