Is this the official Pancyprian Chemistry exam?

No. The official assessment is administered by the Cyprus Examinations Service as a constructed-response, drawing, performance or mixed paper under code 019. This bank is a free English-language MCQ study adaptation for revision; practise every required non-MCQ component with official past papers and the specification.

How is the subject marked?

Like all Pancyprian access subjects it is marked on a 0–20 scale. There is no pass mark; the grade contributes to the access/ranking grade used for admission to public universities in Cyprus and Greece.

What is the 2026 exam fee?

EUR 25.00 per access subject (EUR 50.00 only for Practical Physical Performance Test code 032), payable by card in the official 2–17 April 2026 window, with exemptions for specified groups.

Who can sit this examination?

Final-year student or graduate of a Cyprus public secondary school (lyceum cycle), the English School Nicosia, a licensed private school, or an equivalent foreign upper-secondary school (with certification), within Examinations Service nationality/residence rules. Applications are electronic; un-promoted final-year students with 135+ absences are excluded.

Why is the practice bank in English?

OpenExamPrep publishes one English-language practice URL per exam. Where the tested skill is another language, stems/options may retain that language while explanations stay in English. The bank is not an official translation of the Greek (or other) exam environment.

How should I use this bank with official materials?

Use the bank to diagnose weak syllabus areas, then switch to Θέματα - Λύσεις and the official specification on panexams.moec.gov.cy for timed written, drawing or performance practice that this bank does not simulate.

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100+ Free Pancyprian Chemistry Practice Questions

Cyprus Pancyprian Access Examination Chemistry (ΧΗΜΕΙΑ, code 019) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Pancyprian Chemistry Exam

019

Official subject code for ΧΗΜΕΙΑ in the 2026 Pancyprian Access Examinations

Cyprus Examinations Service

3 hours per the official 2026 timetable

Official examination duration from the 2026 timetable

Cyprus Examinations Service

EUR 25

2026 fee per access subject (not code 032)

Cyprus Examinations Service

0–20

Marking scale for each Pancyprian subject

Cyprus Examinations Service

8–26 June 2026

Examination period of the 2026 Pancyprian Access Examinations

Cyprus Examinations Service

100

Original practice questions in this study bank

OpenExamPrep

Free 100-question English-language MCQ study bank for Pancyprian Chemistry (code 019). Official duration: 3 hours per the official 2026 timetable. Official fee: EUR 25.00 per access subject in 2026. This bank is a study aid, not an official translation or format simulation.

Sample Pancyprian Chemistry Practice Questions

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

1How many moles are contained in 11.0 g of carbon dioxide, CO2? (Molar mass of CO2 = 44.0 g/mol)

A.0.25 mol

B.0.50 mol

C.1.0 mol

D.4.0 mol

Explanation: Moles are calculated as n = m/M = 11.0 g / 44.0 g/mol = 0.25 mol. This is the fundamental mass-to-moles conversion used throughout stoichiometry.

2A solution is prepared by dissolving 0.50 mol of NaCl in water and making the total volume up to 2.0 L. What is the molar concentration of the solution?

A.1.0 M

B.0.50 M

C.0.25 M

D.4.0 M

Explanation: Molarity c = n/V = 0.50 mol / 2.0 L = 0.25 mol/L. Concentration always refers to the final total volume of solution, not the volume of water added.

3How many O2 molecules are present in 2.00 mol of oxygen gas? (Avogadro's constant NA = 6.02 x 10^23 mol^-1)

A.6.02 x 10^23

B.1.204 x 10^24

C.3.01 x 10^23

D.2.41 x 10^24

Explanation: Number of particles N = n x NA = 2.00 mol x 6.02 x 10^23 mol^-1 = 1.204 x 10^24 molecules. One mole of any substance contains Avogadro's number of particles.

450.0 mL of a 2.0 M HCl solution is diluted with water to a final volume of 250 mL. What is the concentration of the diluted solution?

A.0.10 M

B.0.25 M

C.8.0 M

D.0.40 M

Explanation: During dilution the moles of solute stay constant, so c1V1 = c2V2: c2 = (2.0 M x 50.0 mL) / 250 mL = 0.40 M. The volume increased fivefold, so the concentration falls fivefold.

5For the reaction 2H2 + O2 -> 2H2O, a mixture contains 4 mol of H2 and 3 mol of O2. What is the maximum amount of water that can be produced?

A.6 mol

B.4 mol

C.3 mol

D.2 mol

Explanation: 4 mol of H2 requires only 2 mol of O2 (ratio 2:1), so H2 is the limiting reactant. From the 2:2 ratio, 4 mol of H2 produces 4 mol of H2O, leaving 1 mol of O2 in excess.

620 g of NaCl is dissolved in 180 g of water. What is the mass percent (% w/w) concentration of the solution?

A.11.1%

B.20%

C.10%

D.9.1%

Explanation: Mass percent = (mass of solute / mass of solution) x 100 = 20 g / (20 g + 180 g) x 100 = 10%. The denominator is the total mass of the solution, not the mass of the solvent.

7What is the molar mass of calcium hydroxide, Ca(OH)2? (Ar: Ca = 40, O = 16, H = 1)

A.74 g/mol

B.57 g/mol

C.68 g/mol

D.90 g/mol

Explanation: M = 40 + 2 x (16 + 1) = 40 + 34 = 74 g/mol. The subscript 2 outside the bracket multiplies the whole OH group, so the formula unit contains 2 O and 2 H atoms.

8What volume does 1.00 mol of an ideal gas occupy at 273 K and 1.00 atm (standard temperature and pressure)?

A.11.2 L

B.24.5 L

C.44.8 L

D.22.4 L

Explanation: At STP (273 K, 1 atm) one mole of any ideal gas occupies the molar volume 22.4 L. This follows from V = nRT/P with R = 0.0821 L atm K^-1 mol^-1.

9In the ammonia synthesis N2 + 3H2 -> 2NH3, how many moles of NH3 are produced when 9 mol of H2 reacts completely with excess N2?

A.9 mol

B.3 mol

C.6 mol

D.13.5 mol

Explanation: The mole ratio H2:NH3 is 3:2, so n(NH3) = 9 mol x (2/3) = 6 mol. Stoichiometric coefficients give the conversion factor between reactant and product.

10When 50 g of pure CaCO3 is completely decomposed by heating (CaCO3 -> CaO + CO2), what volume of CO2 is released at STP? (M(CaCO3) = 100 g/mol; molar volume at STP = 22.4 L/mol)

A.22.4 L

B.11.2 L

C.5.6 L

D.44.8 L

Explanation: n(CaCO3) = 50 g / 100 g/mol = 0.50 mol. The 1:1 ratio gives 0.50 mol of CO2, which at STP occupies 0.50 x 22.4 L = 11.2 L.

About the Pancyprian Chemistry Practice Questions

Verified exam format metadata for Cyprus Pancyprian Access Examination Chemistry (ΧΗΜΕΙΑ, code 019) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.

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