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100+ Free PSEB SS Chemistry Practice Questions

Punjab School Education Board (PSEB) Senior Secondary Chemistry (Class 12 — Chemistry code 053) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: PSEB SS Chemistry Exam

3 hours

PSEB Senior Secondary Chemistry theory exam duration

PSEB Class 12 Chemistry syllabus 2025–26

70 + 25 + 05

Theory + Practical + INA marks for Class 12 Chemistry (total 100)

PSEB Class 12 Chemistry syllabus

33% each + aggregate

Pass floor in theory, practical, and INA separately plus aggregate

PSEB Scheme of Studies Class XII

Code 053

Chemistry subject code on PSEB Senior Secondary scheme

PSEB Class 12 scheme / syllabus

English MCQ adaptation

This free local bank is not the official Senior Secondary paper format

OpenExamPrep practice policy

PSEB Senior Secondary Chemistry (code 053) is a Class 12 public-exam subject with Theory 70 + Practical 25 + INA 05 (3 hours theory) and 33% component + aggregate pass — not a pure MCQ board paper. This free 2026 bank is an English MCQ study adaptation for NCERT/PSEB-aligned Class 12 Chemistry concepts and calculations.

Sample PSEB SS Chemistry Practice Questions

Try these sample questions to test your PSEB SS Chemistry exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Molarity (M) of a solution is defined as the number of moles of solute present in:
A.One litre of solution
B.One kilogram of solvent
C.One litre of solvent
D.One kilogram of solution
Explanation: Molarity M = moles of solute / volume of solution in litres. Molality uses mass of solvent in kilograms.
2Which concentration unit is independent of temperature?
A.Molarity
B.Normality
C.Molality
D.Volume percentage
Explanation: Molality depends only on moles of solute and mass of solvent. Mass does not change with temperature, so molality is temperature-independent. Molarity and volume-based units change as volume expands or contracts with temperature.
311.7 g of NaCl (M = 58.5 g mol⁻¹) is dissolved in water to make 250 mL of solution. The molarity is:
A.0.8 M
B.0.4 M
C.1.0 M
D.2.0 M
Explanation: Moles of NaCl = 11.7/58.5 = 0.20 mol. Volume = 0.250 L. M = 0.20/0.250 = 0.80 M.
436 g of glucose (M = 180 g mol⁻¹) is dissolved in 500 g of water. The molality of the solution is:
A.0.4 m
B.0.2 m
C.0.5 m
D.0.8 m
Explanation: Moles of glucose = 36/180 = 0.20 mol. Mass of solvent = 0.500 kg. Molality m = 0.20/0.500 = 0.40 m.
5According to Raoult’s law for a volatile solvent A, its partial vapour pressure in an ideal solution is:
A.P°_A × mole fraction of A in the solution
B.Equal to the pure solvent vapour pressure P°_A regardless of composition
C.P°_A × mole fraction of solute only
D.Independent of temperature
Explanation: Raoult’s law: p_A = P°_A · x_A, where x_A is the mole fraction of solvent A in the liquid phase for an ideal solution.
6Colligative properties of dilute solutions depend primarily on the:
A.Number of solute particles relative to solvent particles
B.Chemical identity of the solute only
C.Colour of the solution
D.Density of pure solvent alone
Explanation: Colligative properties (relative vapour pressure lowering, ΔT_b, ΔT_f, osmotic pressure) depend on the relative number of solute particles, not on their chemical identity (for ideal dilute non-associating/non-dissociating cases, adjusted by i when needed).
7A 0.20 m aqueous solution of a non-volatile non-electrolyte has K_b = 0.52 K kg mol⁻¹. The elevation in boiling point is:
A.0.052 K
B.0.520 K
C.0.104 K
D.1.04 K
Explanation: ΔT_b = K_b · m · i. For non-electrolyte i = 1, so ΔT_b = 0.52 × 0.20 = 0.104 K.
8For a 0.10 m NaCl solution assuming complete dissociation (i = 2) and K_f = 1.86 K kg mol⁻¹, the depression in freezing point is:
A.0.093 K
B.0.186 K
C.0.372 K
D.1.86 K
Explanation: ΔT_f = K_f · m · i = 1.86 × 0.10 × 2 = 0.372 K.
9The osmotic pressure (π) of an ideal dilute solution is given by:
A.π = nRT (for pure solvent)
B.π = mRT only for solids
C.π = CRT (C = molar concentration of solute)
D.π = ΔT_b / K_b
Explanation: van’t Hoff equation: π = CRT, where C is molar concentration of solute particles (with i if needed), R gas constant, T absolute temperature.
10At 300 K, the osmotic pressure of a solution containing 2.0 g of a non-electrolyte solute (M = 60 g mol⁻¹) in 500 mL solution is approximately (R = 0.0821 L atm mol⁻¹ K⁻¹):
A.0.82 atm
B.2.46 atm
C.1.64 atm
D.3.28 atm
Explanation: n = 2.0/60 = 1/30 mol. V = 0.500 L. C = n/V = (1/30)/0.5 = 1/15 M. π = CRT = (1/15)×0.0821×300 ≈ 1.642 atm ≈ 1.64 atm.

About the PSEB SS Chemistry Practice Questions

Verified exam format metadata for Punjab School Education Board (PSEB) Senior Secondary Chemistry (Class 12 — Chemistry code 053) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.