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100+ Free RBSE SS Chemistry Practice Questions

Rajasthan RBSE Senior Secondary (Class 12) Chemistry — Code 041 practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: RBSE SS Chemistry Exam

041

RBSE subject code for Chemistry

RBSE 2026 subject-wise statistics

Theory 56 + Sessional 14

Typical full marks / assessment pattern

RBSE syllabus examination scheme

3 h 15 m

Typical theory paper duration (where applicable)

RBSE Class 12 / Praveshika schemes

33%

Common minimum pass threshold under RBSE regulations

RBSE examination regulations (confirm circular)

2026

Main examination cycle evidenced in official subject-wise statistics

rajeduboard.rajasthan.gov.in/statistics2026.htm

MCQ study aid

Local bank adapts knowledge to four-option MCQs; official paper is mixed/performance format

OpenExamPrep assessment-format policy

RBSE Senior Secondary (Class 12) Chemistry (code 041): Theory 56 + Sessional 14 = 70 and Practical 30 = 100; 3 hours 15 minutes (theory). Pass about 33% per RBSE rules (confirm circular). Fee as per board notification. Free English MCQ study aid — not a full official-format simulation.

Sample RBSE SS Chemistry Practice Questions

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

1A solution is prepared by dissolving 0.50 mol of urea (non-volatile, non-electrolyte) in 500 g of water. What is the molality of urea in the solution?
A.0.50 m
B.1.0 m
C.2.0 m
D.0.25 m
Explanation: Molality m = moles of solute / kg of solvent = 0.50 mol / 0.500 kg = 1.0 m. Mass of water is 500 g = 0.500 kg.
2According to Raoult’s law for an ideal binary solution of volatile liquids A and B, the partial vapour pressure of component A is equal to
A.its pure vapour pressure multiplied by its mole fraction in the vapour phase
B.its pure vapour pressure multiplied by its mole fraction in the liquid phase
C.the total pressure multiplied by its mole fraction in the liquid phase
D.its pure vapour pressure divided by its mole fraction in the liquid phase
Explanation: Raoult’s law: p_A = p_A° · x_A, where x_A is the mole fraction of A in the liquid solution and p_A° is the vapour pressure of pure A.
3Which colligative property is preferred for determining the molar mass of a polymer of high molar mass?
A.Relative lowering of vapour pressure
B.Elevation of boiling point
C.Depression of freezing point
D.Osmotic pressure
Explanation: Osmotic pressure (π = CRT) gives measurable effects even for very dilute solutions of high-molar-mass solutes, so it is the preferred colligative method for polymers.
4A 0.10 m aqueous solution of a non-volatile non-electrolyte freezes at −0.186 °C. If K_f for water is 1.86 K kg mol⁻¹, the depression of freezing point confirms that
A.ΔT_f = K_f · m = 0.186 K
B.ΔT_f = K_f / m = 18.6 K
C.ΔT_f = K_f · m² = 0.0186 K
D.ΔT_f is independent of molality for non-electrolytes
Explanation: For a non-electrolyte, ΔT_f = K_f · m = 1.86 × 0.10 = 0.186 K, so the freezing point is −0.186 °C for water.
5Henry’s law states that the solubility of a gas in a liquid at constant temperature is directly proportional to
A.the square of the partial pressure of the gas
B.the partial pressure of the gas above the solution
C.the total pressure of the system only
D.the molar mass of the gas
Explanation: Henry’s law: p = K_H · x (or c = k · p). Solubility of a sparingly soluble gas is proportional to its partial pressure above the liquid at fixed T.
6For a dilute aqueous solution of NaCl (strong electrolyte), the van’t Hoff factor i approaches
A.1
B.2
C.3
D.0.5
Explanation: Complete dissociation of NaCl → Na⁺ + Cl⁻ gives two particles per formula unit, so i ≈ 2 for dilute solutions.
7An ideal solution of two volatile liquids is one in which
A.Δ_mix H = 0 and Δ_mix V = 0
B.Δ_mix H > 0 and Δ_mix V > 0
C.Δ_mix H < 0 and Δ_mix V < 0
D.Δ_mix G = 0
Explanation: For an ideal solution, intermolecular forces A–B equal A–A and B–B averages, so enthalpy and volume of mixing are zero while Δ_mix G is negative (entropy-driven).
8Aqueous glucose (C₆H₁₂O₆) solution of concentration 0.10 mol L⁻¹ at 300 K has osmotic pressure closest to (R = 0.0821 L atm mol⁻¹ K⁻¹)
A.0.82 atm
B.2.46 atm
C.0.246 atm
D.24.6 atm
Explanation: π = CRT = 0.10 × 0.0821 × 300 ≈ 2.463 atm ≈ 2.46 atm for a non-electrolyte.
9In a galvanic cell, the standard electrode potential of Zn²⁺/Zn is −0.76 V and of Cu²⁺/Cu is +0.34 V. The standard EMF of the cell Zn | Zn²⁺ || Cu²⁺ | Cu is
A.1.10 V
B.0.42 V
C.−1.10 V
D.0.34 V
Explanation: E°_cell = E°_cathode − E°_anode = E°_Cu − E°_Zn = 0.34 − (−0.76) = 1.10 V. Copper is reduced (cathode); zinc is oxidised (anode).
10For the half-reaction Zn²⁺ + 2 e⁻ → Zn at 25 °C, the Nernst equation is
A.E = E° − (0.059/2) log [Zn²⁺]
B.E = E° − (0.059/2) log (1/[Zn²⁺])
C.E = E° − (0.059) log (1/[Zn²⁺])
D.E = E° + (0.059/2) log (1/[Zn²⁺])
Explanation: Q for the reduction is 1/[Zn²⁺] (solid Zn activity = 1). Thus E = E° − (0.059/n) log Q with n = 2, so E = E° − (0.059/2) log (1/[Zn²⁺]). Equivalently E = E° + (0.059/2) log [Zn²⁺].

About the RBSE SS Chemistry Practice Questions

Verified exam format metadata for Rajasthan RBSE Senior Secondary (Class 12) Chemistry — Code 041 is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.