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Sample ÖABT Kimya/Kimya Teknolojisi Practice Questions

Try these sample questions to review concepts for the ÖABT Kimya/Kimya Teknolojisi exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What is the maximum number of electrons in an atom that can simultaneously possess the quantum numbers n = 4 and l = 2?

A.2

B.6

C.10

D.14

Explanation: The quantum numbers n = 4 and l = 2 define the 4d subshell. A d subshell has 5 orbitals (ml = -2, -1, 0, +1, +2), and each orbital can hold a maximum of 2 electrons with opposite spins, yielding a total capacity of 10 electrons.

2Complete combustion of 0.100 mol of an unknown hydrocarbon CxHy produces exactly 0.300 mol of CO2(g) and 0.400 mol of H2O(g). What is the molecular formula of this hydrocarbon?

A.C3H6

B.C3H8

C.C4H8

D.C4H10

Explanation: From 0.100 mol of CxHy, 0.300 mol of CO2 is formed, which means each mole of hydrocarbon contains 0.300 / 0.100 = 3 moles of carbon atoms (x = 3). Similarly, 0.400 mol of H2O contains 0.800 mol of hydrogen atoms, meaning each mole of hydrocarbon contains 0.800 / 0.100 = 8 moles of hydrogen atoms (y = 8). Thus, the molecular formula is C3H8 (propane).

3A sample of an ideal gas occupies 2.00 L at 1.00 atm and 300 K. If the gas is compressed to a volume of 0.500 L while the temperature is increased to 450 K, what is the final pressure?

A.2.67 atm

B.4.00 atm

C.6.00 atm

D.8.00 atm

Explanation: Applying the combined gas law, (P1 * V1) / T1 = (P2 * V2) / T2. Rearranging for P2 gives P2 = P1 * (V1 / V2) * (T2 / T1) = 1.00 atm * (2.00 L / 0.500 L) * (450 K / 300 K) = 1.00 * 4.00 * 1.50 = 6.00 atm.

4Under identical temperature and pressure conditions, an unknown gaseous compound X effuses through a pinhole at 0.500 times the rate of oxygen gas, O2 (molar mass = 32.0 g/mol). What is the molar mass of compound X?

A.16.0 g/mol

B.64.0 g/mol

C.96.0 g/mol

D.128 g/mol

Explanation: According to Graham's law of effusion, Rate(X) / Rate(O2) = sqrt(M(O2) / M(X)). Substituting the given values gives 0.500 = sqrt(32.0 / M(X)). Squaring both sides yields 0.250 = 32.0 / M(X), so M(X) = 32.0 / 0.250 = 128 g/mol.

5Which of the following correctly explains why the first ionization energy of phosphorus (15P) is higher than that of sulfur (16S), despite sulfur having a higher nuclear charge?

A.Phosphorus has a significantly smaller atomic radius than sulfur.

B.Phosphorus has a stable, half-filled 3p subshell, whereas sulfur experiences inter-electronic repulsion in a doubly occupied 3p orbital.

C.Sulfur's valence electrons experience greater effective nuclear charge shielding from inner d electrons.

D.Sulfur easily loses two electrons simultaneously to achieve a noble gas configuration.

Explanation: The valence electron configuration of P is [Ne] 3s2 3p3, where the 3p subshell is half-filled with parallel spins, imparting exchange energy stability. Sulfur has [Ne] 3s2 3p4, in which one 3p orbital contains a paired set of electrons; the electrostatic repulsion between these two paired electrons makes removing one electron energetically easier than removing an electron from phosphorus.

6A binary compound consists of 4.00 g of calcium (atomic mass = 40.0 g/mol) and 16.0 g of bromine (atomic mass = 80.0 g/mol). What is the empirical formula of this compound?

A.CaBr

B.CaBr2

C.Ca2Br

D.CaBr3

Explanation: Calculate the moles of each constituent element: Moles of Ca = 4.00 g / 40.0 g/mol = 0.100 mol. Moles of Br = 16.0 g / 80.0 g/mol = 0.200 mol. Dividing by the smaller quantity yields a molar ratio of Ca : Br = 0.100 / 0.100 : 0.200 / 0.100 = 1 : 2. Therefore, the empirical formula is CaBr2.

7In the thermite reaction, 5.40 g of aluminum powder (molar mass = 27.0 g/mol) is mixed with 32.0 g of iron(III) oxide, Fe2O3 (molar mass = 160 g/mol): 2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s). Assuming the reaction proceeds to completion, what mass of iron metal (molar mass = 55.85 g/mol) is produced?

A.5.59 g

B.11.2 g

C.22.3 g

D.32.0 g

Explanation: Moles of Al = 5.40 g / 27.0 g/mol = 0.200 mol. Moles of Fe2O3 = 32.0 g / 160 g/mol = 0.200 mol. By stoichiometry, 2 mol of Al requires 1 mol of Fe2O3, so 0.200 mol of Al requires 0.100 mol of Fe2O3. Thus, Al is the limiting reactant. From 0.200 mol of Al, 0.200 mol of Fe is produced (1:1 mole ratio). Mass of Fe = 0.200 mol * 55.85 g/mol = 11.17 g, which rounds to 11.2 g.

8A solution is prepared by dissolving 18.0 g of glucose (C6H12O6, molar mass = 180 g/mol) in 250 g of pure water. What is the molality (m) of the resulting glucose solution?

A.0.100 m

B.0.250 m

C.0.400 m

D.0.720 m

Explanation: Molality (m) is defined as moles of solute per kilogram of solvent. Moles of glucose = 18.0 g / 180 g/mol = 0.100 mol. Mass of water solvent = 250 g = 0.250 kg. Therefore, molality = 0.100 mol / 0.250 kg = 0.400 m (mol/kg).

9What is the freezing point of an aqueous solution containing 5.85 g of NaCl (molar mass = 58.5 g/mol) dissolved in 500 g of water, assuming complete ionic dissociation? (Kf for water = 1.86 °C*kg/mol)

A.-0.186 °C

B.-0.372 °C

C.-0.744 °C

D.-1.488 °C

Explanation: Moles of NaCl = 5.85 g / 58.5 g/mol = 0.100 mol. Mass of solvent = 0.500 kg, giving a molality m = 0.100 mol / 0.500 kg = 0.200 m. Since NaCl dissociates into Na+ and Cl- ions, the van 't Hoff factor i = 2. Freezing point depression Delta Tf = i * Kf * m = 2 * 1.86 °C*kg/mol * 0.200 mol/kg = 0.744 °C. Since pure water freezes at 0.00 °C, the freezing point is -0.744 °C.

10A 1.00 g sample of an unknown non-electrolyte protein is dissolved in sufficient water to make 100 mL of solution. At 300 K, this solution exhibits an osmotic pressure of 0.0246 atm. Given the gas constant R = 0.08206 L*atm/(mol*K), what is the molar mass of the protein?

A.1,000 g/mol

B.5,000 g/mol

C.10,000 g/mol

D.20,000 g/mol

Explanation: Osmotic pressure is given by Pi = M * R * T, where M is molarity. M = Pi / (R * T) = 0.0246 atm / (0.08206 L*atm/(mol*K) * 300 K) = 0.0246 / 24.618 = 0.00100 M (mol/L). The number of moles in 100 mL (0.100 L) is n = M * V = 0.00100 mol/L * 0.100 L = 1.00 * 10^-4 mol. The molar mass is mass / moles = 1.00 g / (1.00 * 10^-4 mol) = 10,000 g/mol.

About the ÖABT Kimya/Kimya Teknolojisi Exam

ÖABT Kimya/Kimya Teknolojisi is a separately named ÖSYM field-knowledge assessment used in the MEB Academy selection system. Its published scope includes general, analytical, inorganic, organic and physical chemistry.

Exam sponsor: ÖSYM / Milli Eğitim Bakanlığı (MEB). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

One written field test of 50 five-option MCQs in Turkish, lasting 90 minutes. No mandatory oral, practical, assignment or separate case-study submission occurs within this paper. The mandatory AGS session and subsequent Academy training and assessments are separate.

Time Limit

90 minutes

Passing Score

No universal pass mark. Standardized scores support selection; at least 1 net in AGS and the relevant ÖABT test is required for score calculation.

Exam / Certification Fees

TRY 800 for ÖABT (late application TRY 1,200), plus the mandatory AGS session at TRY 800 (late TRY 1,200).

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

2026 ÖSYM MEB-AGS/ÖABT guide · Source checked 2026-10-21Identity, eligibility, language/format, fees, duration, scoring and Table 1 scope.
2026 results announcement · Source checked 2026-10-21Sitting held 26 July 2026; results released 26 August 2026.
Thermochemistry · Source checked 2026-10-21Content reference for independent study; not an endorsement of this bank.
Coordination chemistry · Source checked 2026-10-21Content reference for independent study; not an endorsement of this bank.
Crystal field theory · Source checked 2026-10-21Content reference for independent study; not an endorsement of this bank.
Organic mechanisms · Source checked 2026-10-21Content reference for independent study; not an endorsement of this bank.

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

32% of official paper

General chemistry

General chemistry. Published approximate weight in the 2026 guide, Table 1.

20% of official paper

Analytical chemistry

Analytical chemistry. Published approximate weight in the 2026 guide, Table 1.

16% of official paper

Inorganic chemistry

Inorganic chemistry. Published approximate weight in the 2026 guide, Table 1.

20% of official paper

Organic chemistry

Organic chemistry. Published approximate weight in the 2026 guide, Table 1.

12% of official paper

Physical chemistry

Physical chemistry. Published approximate weight in the 2026 guide, Table 1.

Preparing for the ÖABT Kimya/Kimya Teknolojisi Exam

What You Need to Know

Passing score: No universal pass mark. Standardized scores support selection; at least 1 net in AGS and the relevant ÖABT test is required for score calculation.
Assessment: One written field test of 50 five-option MCQs in Turkish, lasting 90 minutes. No mandatory oral, practical, assignment or separate case-study submission occurs within this paper. The mandatory AGS session and subsequent Academy training and assessments are separate.
Time limit: 90 minutes
Exam / certification fees: TRY 800 for ÖABT (late application TRY 1,200), plus the mandatory AGS session at TRY 800 (late TRY 1,200). Official sources

Using Our Practice Resources

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ÖABT Kimya/Kimya Teknolojisi: Suggested Study Strategy

1Use the 2026 guide and applicable MEB Academy notice to check the official field and current application conditions.

2Review the Turkish terminology and primary subject resources alongside these English explanations.

3Complete additional Turkish-language study across the published outline; this selected set does not exhaust the field.

Frequently Asked Questions

What are the eligibility and other assessment requirements?

The 2026 guide requires the statutory Academy-entry conditions. Graduation from a MEB-approved higher-education program for the intended field is required by the Academy application date. Check the applicable MEB recruitment notice for accepted programs and other conditions; test participation does not confer an appointment right. All candidates must take the morning AGS session; subsequent Academy training is separate.

How is the real examination scored?

Raw score is correct answers minus one-quarter of wrong answers. Scores are standardized and combined under the guide’s field-specific rules; there is no universal pass mark or automatic appointment.