Entrance exam in medicine
Chemistry Test
The chemistry test is based on the compulsory chemistry courses of the Estonian curriculum of upper secondary school.
- The development of chemistry as a science. Physical and chemical research methods in Chemistry. Chemistry-related career-choices.
- Structure of matter. Modern understanding of the structure of atom. Information in the periodic table and interpreting it. Types of chemical bonds. Hydrogen bond. Intermolecular forces. The relationship between the physical properties of a matter and the structure of a matter.
- Why and how do chemical reactions occur? Reaction activation energy, active collisions. Exothermic and endothermic reactions. The rate of a chemical reaction and factors influencing the reaction rate. Chemical equilibrium and its shifting (the Le Chatelier principle).
- Dissolution process, chemical reactions in solutions. Dissolution process of substances. Electrolytes and non-electrolytes; strong and weak electrolytes. Protolytic theory of acids and bases. Molar concentration. Interionic reactions in solutions, the conditions of their course. pH. Environment (pH) in the solution of a hydrolysing salt.
- Metals. Overview of the characteristic physical and chemical properties of metals. Comparison of the chemical activity of metals; activity series of metals. Metals and their compounds in everyday life and nature. Redox processes related to metals: producing metals from ore, electrolysis, corrosion, galvanic cells (without requiring reaction equations). Taking into account reaction yield and additivities in molar calculations based on a reaction equation.
- Non-metals. Overview of the physical and chemical properties of non-metals depending on the position of the element in the periodic table. Comparison of the chemical activity of non-metals. Overview of some non-metals and their compounds (chosen freely, on the example of natural and/or industrial processes).
- Hydrocarbons and their derivates. Structure of carbon compounds and the ways of describing it. Alkanes, nomenclature principles, isomerism. Dependence of physical properties of substituted alkanes (halogen compounds, alcohols, primary amines) on their structure. Comparison of the chemical properties of unsaturated and aromatic hydrocarbons and alkanes. Chain-growth polymerisation. Hydrocarbons and their derivates in nature and industry.
- Organic substances around us. Aldehydes as the oxidation products of alcohols. Substituted carboxylic acids (amino acids, hydroxy acids) and functional derivates of carboxylic acids (esters, amides). Polycondensation. Organic compounds in living organisms: fats, saccharides, proteins.
Curricula for
- upper secondary school (courses: „Fundamentals of Chemistry“, „Inorganic substances“, and „ Organic compounds“) and
- basic school.
The chemistry test contains different types of questions and tasks. The applicants can use the periodic table, activity series of metals and the solubility table (the applicant can use them in the test environment).
The sample exercises give an idea of the format in which questions will appear in the exam. The exam questions are based on four main topics:
- general chemistry,
- inorganic chemistry,
- organic chemistry, and
- calculation-based tasks.
Each topic makes up approximately one-quarter of the entire Chemistry test.
Please find here a Chemistry sample test.
Please find here a Periodic table, activity series of metals and the solubility table.
The atomic masses of the elements in the attached periodic table of elements must be used in the calculations. For example: Ar(H) = 1, Ar(O) = 16 and Ar(Cl) = 35.5
Note! Some exercises contain multiple sub-tasks, which are placed on separate pages.
SAMPLE EXERCISES
Exercise 1
Exercise 2A
Exercise 2B
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Select the appropriate name of the compound, structural formula and/or class of organic compound that corresponds to the given source information, which is missing from the data. Determine whether hydrogen bonding occurs in this substance.
Click on the correct answers.
Exercise 7
Exercise 8
Exercise 9
A. Calculate the amount of oxygen atoms (in moles)
a) In 44.8 dm³ of nitrogen dioxide gas (STP: 1 atm and 0 °C).
n(NO2) = V /Vm = 44.8 dm3/(22.4 dm3/mol) = 2 mol
1 mol NO2 ⇔ 2 mol O
n(O) = 2n(NO2) = 2 ⋅ 2 mol = 4 mol
Answer: 44.8 dm³ of nitrogen dioxide contains 4 moles of oxygen atoms.
b) In 426 g of aluminium nitrate.
M(Al(NO3)3) = 213 g/mol
n(Al(NO3)3) = m /M = 426 g/(213 g/mol) = 2 mol
1 mol Al(NO3)3 ⇔ 9 mol O
n(O) = 9n(Al(NO3)3) = 9 ⋅ 2 mol = 18 mol
Answer: 426 g of aluminium nitrate contains 18 moles of oxygen atoms.
B. Which contains a smaller amount of oxygen: 1 mole of nitrogen dioxide or 1 mole of aluminium nitrate?
Answer: 1 mole of nitrogen dioxide contains smaller amount of oxygen.
Justification:
NO2: 1 mol NO2 ⇔ 2 mol O
Al(NO3)3: 1 mol Al(NO3)3 ⇔ 9 mol O
2 mol O < 9 mol O
Exercise 10
A. Calculate how many moles of ethylene glycol are contained in 1.00 dm³ of fresh antifreeze (i.e. find the molar concentration of a 35.0% ethylene glycol solution). The dependence of the solution’s density on mass percent is shown in the graph.
ρ(35% solution) = 1.043 g/cm3
m(35% solution) = m ∙ V = 1.00 dm3 ∙ 1000 cm3/1 dm3 ∙ 1.043 g/cm3 = 1043 g
m(ethylene glycol) = 0.350 ∙ 1043 g = 365 g
M(CH2(OH)CH2(OH)) = 62 g/mol
n(ethylene glycol) = m /M = 365 g/(62 g/mol) = 5.89 mol
c(ethylene glycol) = n /V = 5,89 mol/1,00 dm3 = 5.89 mol/dm3
Answer: 1.00 dm³ of fresh antifreeze contains 5,89 moles of ethylene glycol.
B. During use, 120 cm³ of water evaporated from the coolant described in part A. Calculate the mass percent of ethylene glycol in the coolant after partial evaporation of water.
m(evaporated water) = 120 g (ρ(vesi) = 1,00 g/cm3)
m(used coolant) = 1043 g – 120 g = 923 g
%(ethylene glycol in used coolant) = 365 g/923 g ∙ 100 = 39.5
Answer: The used coolant contains 39.5 mass percent of ethylene glycol.
Exercise 11
A. a) Calculate how many moles of phosphoric acid were produced in the reaction if 12.0% of the P₄O₁₀ did not react.
P4O10 + 6 H2O = 4 H3PO4
Identify the reagent in excess.
1 mol P4O10 ⇔ 6 mol H2O
n(H2O) = 6/1 ∙ 15 mol = 90 mol < 100 mol Water is in excess.
1 mol P4O10 ⇔ 4 mol H3PO4
n(P4O10 reacted) = 15.0 mol ∙ (1 – 0.120) = 13.2 mol
n(H3PO4) = 4/1 ∙ 13.2 mol = 52.8 mol
Answer: The amount of product formed was 52.8 moles.
b) Calculate the reaction yield percentage based on the initial amount of phosphorus(V) oxide.
Yield = 100% − 12.0% = 88.0%
Answer: The reaction yield was 88.0 %.
B. Calculate how many moles of water remained unreacted.
n(water reacted) = 6/1 ∙ 13.2 mol = 79.2 mol
n(water unreacted) = 100.0 mol − 79.2 mol = 20.8 mol
Answer: The amount of unreacted water was 20.8 moles.
