Physical Chemistry Mock Tests

331 questions available

Physical Chemistry Mock Test 1

Questions: 30

Physical Chemistry Mock Test 2

Questions: 30

Physical Chemistry Mock Test 3

Questions: 30

Physical Chemistry Mock Test 4

Questions: 30

Physical Chemistry Mock Test 5

Questions: 30

Physical Chemistry Mock Test 6

Questions: 30

Physical Chemistry Mock Test 7

Questions: 30

Physical Chemistry Mock Test 8

Questions: 30

Physical Chemistry Mock Test 9

Questions: 30

Physical Chemistry Mock Test 10

Questions: 30

Physical Chemistry Mock Test 11

Questions: 30

Physical Chemistry Mock Test 12

Questions: 1

Sample Questions

JEE Advanced Chemistry

The lattice parameter of a cubic crystal with density 2.70 g·cm⁻³, molar mass 26.98 g/mol, and unit cell edge length 405 pm is:

A Hexagonal close-packed

B Face-centered cubic (fcc)

C Simple cubic

D Body-centered cubic (bcc)

AEEE Chemistry

The number of atoms per unit cell in a face-centered cubic (fcc) lattice is:

A 1

B 2

C 3

D 4

BITSAT Chemistry

In the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −198 kJ, which condition favours the formation of SO₃?

A High temperature and high pressure

B Low temperature and low pressure

C High temperature and low pressure

D Low temperature and high pressure

BITSAT Chemistry

The van't Hoff factor (i) for a 0.1 M Ba(NO₃)₂ solution is 2.74. The degree of dissociation of Ba(NO₃)₂ is:

A 91.3%

B 74%

C 87%

D 50%

BITSAT Chemistry

The rate of a reaction doubles when temperature increases from 300 K to 310 K. The activation energy of the reaction is: (R = 8.314 J/(mol·K))

A 53.6 kJ/mol

B 45.2 kJ/mol

C 68.3 kJ/mol

D 38.7 kJ/mol

VITEEE Chemistry

At equilibrium, the relationship between ΔG° and the equilibrium constant K is:

A ΔG° = −RT ln K

B ΔG° = RT ln K

C ΔG° = −RT/K

D ΔG° = RT/K

BITSAT Chemistry

The pH of a solution with [H⁺] = 2.5 × 10⁻⁴ M is:

A 3.80

B 4.20

C 3.40

D 3.60

VITEEE Chemistry

The standard Gibbs free energy change (ΔG⁰) for a reaction is related to the equilibrium constant K by:

A ΔG⁰ = −RT ln K

B ΔG⁰ = RT ln K

C ΔG⁰ = −RT/K

D ΔG⁰ = RT/K

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