75 free MCQs on Some Basic Concepts of Chemistry with worked answers and explanations. The mole is chemistry's counting unit. Learn to relate mass, volume, and number of particles using Avogadro's number and molar mass, the foundation of all stoichiometry and reaction calculations.
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Mass connects to moles via molar mass, and moles connect to particle count via Avogadro's number or to gas volume via the 22.4 L/mol rule at STP.
Q1.
The law of conservation of mass, which states that mass can neither be created nor destroyed in a chemical reaction, was proposed by:
Answer: A. Antoine Lavoisier
Why: Antoine Lavoisier established the law of conservation of mass: the total mass of reactants equals the total mass of products in a chemical change.
Q2.
According to the law of definite proportions, a given chemical compound always contains its constituent elements combined in a fixed proportion by:
Answer: B. mass
Why: Proust's law of definite (constant) proportions states that a pure compound always has the same elements combined in the same fixed ratio by mass.
Q3.
Which of the following is an intensive property that does not depend on the amount of substance present?
Answer: C. Temperature
Why: Temperature is an intensive property; its value is independent of the quantity of matter. Mass, volume and number of moles are extensive properties.
Q4.
Avogadro's law states that equal volumes of all gases under the same temperature and pressure contain an equal number of:
Answer: D. molecules
Why: Avogadro's law: under identical conditions of temperature and pressure, equal volumes of all gases contain an equal number of molecules.
Q5.
The sum of the mole fractions of all the components present in a solution is always equal to:
Answer: A. one
Why: Mole fraction is the ratio of moles of a component to the total moles; summing them over all components always gives exactly 1.
Q6.
What is one mole of any substance equal to?
Answer: A. 6.022 x 10<sup>23</sup> particles
Why: One mole contains Avogadro's number of particles, which is 6.022 x 10<sup>23</sup> particles.
Q7.
What is Avogadro's number?
Answer: A. 6.022 x 10<sup>23</sup>
Why: Avogadro's number is 6.022 x 10<sup>23</sup>, the number of particles present in one mole.
Q8.
The SI unit for amount of substance is:
Answer: B. Mole
Why: The mole is the SI unit used to measure the amount of substance.
Q9.
How many molecules are present in 1 mole of water?
Answer: A. 6.022 x 10<sup>23</sup>
Why: One mole of any molecular substance contains 6.022 x 10<sup>23</sup> molecules.
Q10.
What is the molar mass of water, H<sub>2</sub>O?
Answer: B. 18 g/mol
Why: H<sub>2</sub>O has 2 hydrogen atoms and 1 oxygen atom, so molar mass = 2(1) + 16 = 18 g/mol.
Q11.
How many moles are present in 18 g of water?
Answer: B. 1 mol
Why: Moles = mass / molar mass. For water, 18 g / 18 g/mol = 1 mol.
Q12.
What formula is used to calculate number of moles from mass?
Answer: B. Moles = Mass / Molar mass
Why: The correct formula is moles = given mass divided by molar mass.
Q13.
What is the molar mass of carbon dioxide, CO<sub>2</sub>?
Answer: C. 44 g/mol
Why: CO<sub>2</sub> has 1 carbon atom and 2 oxygen atoms, so molar mass = 12 + 2(16) = 44 g/mol.
Q14.
How many moles are present in 44 g of CO<sub>2</sub>?
Answer: B. 1 mol
Why: Moles = mass / molar mass. CO<sub>2</sub> has molar mass 44 g/mol, so 44 g is 1 mole.
Q15.
At STP, 1 mole of any gas occupies:
Answer: C. 22.4 L
Why: At STP, one mole of any ideal gas occupies 22.4 litres.
Q16.
How many litres are occupied by 2 moles of a gas at STP?
Answer: C. 44.8 L
Why: At STP, 1 mole of gas occupies 22.4 L, so 2 moles occupy 44.8 L.
Q17.
The molar mass of oxygen gas, O<sub>2</sub>, is:
Answer: C. 32 g/mol
Why: One oxygen atom has atomic mass 16, so O<sub>2</sub> has molar mass 2 x 16 = 32 g/mol.
Q18.
How many moles are present in 32 g of oxygen gas, O<sub>2</sub>?
Answer: B. 1 mol
Why: O<sub>2</sub> has molar mass 32 g/mol, so 32 g of O<sub>2</sub> equals 1 mole.
Q19.
What is the molar mass of NaCl?
Answer: C. 58.5 g/mol
Why: Na has atomic mass 23 and Cl has atomic mass 35.5, so NaCl = 23 + 35.5 = 58.5 g/mol.
Q20.
How many moles are present in 58.5 g of NaCl?
Answer: B. 1 mol
Why: NaCl has molar mass 58.5 g/mol, so 58.5 g of NaCl equals 1 mole.
Q21.
Which quantity connects mass and number of particles?
Answer: A. Mole
Why: The mole connects the mass of a substance with the number of particles it contains.
Q22.
How many atoms are present in 1 mole of helium?
Answer: C. 6.022 x 10<sup>23</sup> atoms
Why: Helium is monoatomic, and 1 mole of helium contains 6.022 x 10<sup>23</sup> helium atoms.
Q23.
What is the molar mass of methane, CH<sub>4</sub>?
Answer: B. 16 g/mol
Why: CH<sub>4</sub> has 1 carbon atom and 4 hydrogen atoms, so molar mass = 12 + 4(1) = 16 g/mol.
Q24.
How many moles are present in 16 g of methane, CH<sub>4</sub>?
Answer: B. 1 mol
Why: Methane has molar mass 16 g/mol, so 16 g of CH<sub>4</sub> equals 1 mole.
Q25.
In a balanced chemical equation, coefficients represent the ratio of:
Answer: A. Moles
Why: Coefficients in a balanced equation show the mole ratio between reactants and products.
Q26.
Chlorine occurs as two isotopes, <sup>35</sup>Cl (75%) and <sup>37</sup>Cl (25%). Its average atomic mass is:
Answer: B. 35.5 u
Why: Average atomic mass = (35 x 0.75) + (37 x 0.25) = 26.25 + 9.25 = 35.5 u.
Q27.
Carbon forms two oxides, CO and CO<sub>2</sub>. For a fixed mass of carbon, the masses of oxygen that combine are in the simple whole-number ratio:
Answer: D. 1:2
Why: Per 12 g of carbon, CO has 16 g oxygen and CO<sub>2</sub> has 32 g oxygen; the ratio 16:32 = 1:2, illustrating the law of multiple proportions.
Q28.
What is the molality of a solution prepared by dissolving 8 g of NaOH (molar mass 40 g mol<sup>-1</sup>) in 500 g of water?
Answer: A. 0.4 m
Why: Moles of NaOH = 8/40 = 0.2 mol; mass of solvent = 0.5 kg; molality = 0.2/0.5 = 0.4 mol kg<sup>-1</sup>.
Q29.
A solution contains 1 mole of water and 4 moles of ethanol. The mole fraction of ethanol in the solution is:
Answer: C. 0.8
Why: Mole fraction of ethanol = 4/(1+4) = 4/5 = 0.8.
Q30.
What volume of water must be added to 100 mL of 0.5 M HCl to dilute it to 0.1 M?
Answer: B. 400 mL
Why: Using M1V1 = M2V2: 0.5 x 100 = 0.1 x V2, so V2 = 500 mL total. Water to be added = 500 - 100 = 400 mL.
Q31.
How many moles are present in 90 g of water?
Answer: C. 5 mol
Why: Molar mass of water is 18 g/mol. Moles = 90/18 = 5 mol.
Q32.
How many molecules are present in 2 moles of CO<sub>2</sub>?
Answer: B. 1.2044 x 10<sup>24</sup>
Why: Number of molecules = moles x Avogadro's number = 2 x 6.022 x 10<sup>23</sup> = 1.2044 x 10<sup>24.</sup>
Q33.
How many moles are present in 11.2 L of a gas at STP?
Answer: B. 0.5 mol
Why: At STP, 1 mole of gas occupies 22.4 L. So 11.2 L = 11.2/22.4 = 0.5 mol.
Q34.
What is the mass of 0.5 mole of CO<sub>2</sub>?
Answer: B. 22 g
Why: Molar mass of CO<sub>2</sub> is 44 g/mol. Mass = 0.5 x 44 = 22 g.
Q35.
How many atoms are present in 2 moles of helium?
Answer: B. 1.2044 x 10<sup>24</sup>
Why: Helium is monoatomic. Atoms = 2 x 6.022 x 10<sup>23</sup> = 1.2044 x 10<sup>24</sup> atoms.
Q36.
What is the molar mass of calcium carbonate, CaCO<sub>3</sub>?
Answer: C. 100 g/mol
Why: CaCO<sub>3</sub> = 40 + 12 + 3(16) = 100 g/mol.
Q37.
How many moles are present in 100 g of CaCO<sub>3</sub>?
Answer: B. 1 mol
Why: CaCO<sub>3</sub> has molar mass 100 g/mol. Moles = 100/100 = 1 mol.
Q38.
What is the mass of 3 moles of oxygen gas, O<sub>2</sub>?
Answer: D. 96 g
Why: Molar mass of O<sub>2</sub> is 32 g/mol. Mass = 3 x 32 = 96 g.
Q39.
How many moles are present in 3.01 x 10<sup>23</sup> molecules?
Answer: B. 0.5 mol
Why: 3.01 x 10<sup>23</sup> is half of 6.022 x 10<sup>23</sup>, so it is 0.5 mole.
Q40.
How many molecules are present in 9 g of water?
Answer: A. 3.011 x 10<sup>23</sup>
Why: 9 g water = 9/18 = 0.5 mol. Molecules = 0.5 x 6.022 x 10<sup>23</sup> = 3.011 x 10<sup>23.</sup>
Q41.
What volume will 0.25 mole of a gas occupy at STP?
Answer: B. 5.6 L
Why: Volume at STP = moles x 22.4 L = 0.25 x 22.4 = 5.6 L.
Q42.
In the reaction 2H<sub>2</sub> + O<sub>2</sub> -> 2H<sub>2</sub>O, the mole ratio of H<sub>2</sub> to O<sub>2</sub> is:
Answer: C. 2:1
Why: The balanced equation shows 2 moles of H<sub>2</sub> react with 1 mole of O<sub>2</sub>, so the ratio is 2:1.
Q43.
How many moles of H<sub>2</sub>O are formed from 2 moles of H<sub>2</sub> in 2H<sub>2</sub> + O<sub>2</sub> -> 2H<sub>2</sub>O?
Answer: B. 2 mol
Why: The mole ratio H<sub>2</sub>:H<sub>2</sub>O is 2:2, so 2 moles of H<sub>2</sub> form 2 moles of H<sub>2</sub>O.
Q44.
What is the molar mass of sulfuric acid, H<sub>2</sub>SO<sub>4</sub>?
Answer: B. 98 g/mol
Why: H<sub>2</sub>SO<sub>4</sub> = 2(1) + 32 + 4(16) = 98 g/mol.
Q45.
How many moles are present in 49 g of H<sub>2</sub>SO<sub>4</sub>?
Answer: B. 0.5 mol
Why: Molar mass of H<sub>2</sub>SO<sub>4</sub> is 98 g/mol. Moles = 49/98 = 0.5 mol.
Q46.
The number of significant figures in the measurement 0.00206 is:
Answer: A. 3
Why: Leading zeros are not significant, so only 2, 0 and 6 count - three significant figures.
Q47.
How many moles are present in 22 g of carbon dioxide (CO₂)?
Answer: C. 0.5
Why: Molar mass of CO₂ is 44 g/mol, so moles = 22/44 = 0.5 mol.
Q48.
The molarity of a solution containing 0.5 mol of solute in 250 mL of solution is:
Answer: C. 2 M
Why: Molarity = moles / litres = 0.5 / 0.250 = 2 M.
Q49.
Which 1 g sample contains the greatest number of molecules?
Answer: A. 1 g of H₂
Why: H₂ has the smallest molar mass (2 g/mol), so 1 g is the most moles (0.5), hence the most molecules.
Q50.
The percentage by mass of nitrogen in ammonia (NH₃) is:
Answer: A. 82.4%
Why: Mass of N = 14 in a molar mass of 17, so % N = (14/17) × 100 ≈ 82.4%.
Q51.
A compound contains 40% C, 6.7% H and 53.3% O by mass and has a molar mass of 180 g mol<sup>-1</sup>. Its molecular formula is:
Answer: C. C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>
Why: Mole ratio C:H:O = 40/12 : 6.7/1 : 53.3/16 = 3.33 : 6.7 : 3.33 = 1:2:1, giving empirical formula CH<sub>2</sub>O (mass 30). 180/30 = 6, so molecular formula = C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>.
Q52.
For the reaction N<sub>2</sub> + 3H<sub>2</sub> -> 2NH<sub>3</sub>, if 1 mole of N<sub>2</sub> is mixed with 2 moles of H<sub>2</sub>, the limiting reagent is:
Answer: B. H<sub>2</sub>
Why: 1 mole N<sub>2</sub> requires 3 moles H<sub>2</sub>, but only 2 moles are available, so H<sub>2</sub> runs out first and is the limiting reagent.
Q53.
In a reaction the theoretical yield of a product is 25 g, but only 20 g is actually obtained. The percentage yield is:
Answer: A. 80%
Why: Percentage yield = (actual/theoretical) x 100 = (20/25) x 100 = 80%.
Q54.
The molarity of pure water, taking its density as 1 g mL<sup>-1</sup>, is approximately:
Answer: D. 55.5 M
Why: 1 L of water = 1000 g; moles = 1000/18 = 55.5 mol, so the molarity of pure water is about 55.5 M.
Q55.
What volume of O<sub>2</sub> at STP is required for the complete combustion of 2.2 g of propane? (C<sub>3</sub>H<sub>8</sub> + 5O<sub>2</sub> -> 3CO<sub>2</sub> + 4H<sub>2</sub>O)
Answer: D. 5.6 L
Why: Moles of propane = 2.2/44 = 0.05 mol; O<sub>2</sub> needed = 5 x 0.05 = 0.25 mol; volume = 0.25 x 22.4 = 5.6 L at STP.
Q56.
How many grams of CO<sub>2</sub> contain 3.011 x 10<sup>23</sup> molecules?
Answer: B. 22 g
Why: 3.011 x 10<sup>23</sup> molecules = 0.5 mol. Mass of CO<sub>2</sub> = 0.5 x 44 = 22 g.
Q57.
How many oxygen atoms are present in 1 mole of CO<sub>2</sub> molecules?
Answer: B. 1.2044 x 10<sup>24</sup>
Why: Each CO<sub>2</sub> molecule has 2 oxygen atoms, so 1 mole CO<sub>2</sub> contains 2 moles of oxygen atoms = 1.2044 x 10<sup>24</sup> atoms.
Q58.
How many moles of oxygen atoms are present in 2 moles of H<sub>2</sub>SO<sub>4</sub>?
Answer: D. 8 mol
Why: Each H<sub>2</sub>SO<sub>4</sub> molecule has 4 oxygen atoms, so 2 moles H<sub>2</sub>SO<sub>4</sub> contain 8 moles of oxygen atoms.
Q59.
In N<sub>2</sub> + 3H<sub>2</sub> -> 2NH<sub>3</sub>, how many moles of NH<sub>3</sub> form from 1 mole of N<sub>2</sub>?
Answer: B. 2 mol
Why: The balanced equation shows 1 mole N<sub>2</sub> forms 2 moles NH<sub>3</sub>.
Q60.
In N<sub>2</sub> + 3H<sub>2</sub> -> 2NH<sub>3</sub>, how many moles of H<sub>2</sub> are needed for 2 moles of N<sub>2</sub>?
Answer: C. 6 mol
Why: The ratio N<sub>2</sub>:H<sub>2</sub> is 1:3. For 2 moles of N<sub>2</sub>, hydrogen needed = 2 x 3 = 6 moles.
Q61.
What mass of CaCO<sub>3</sub> is needed to produce 44 g of CO<sub>2</sub> in CaCO<sub>3</sub> -> CaO + CO<sub>2</sub>?
Answer: B. 100 g
Why: 44 g CO<sub>2</sub> is 1 mole. The equation ratio CaCO<sub>3</sub>:CO<sub>2</sub> is 1:1, so 1 mole CaCO<sub>3</sub> = 100 g is needed.
Q62.
How many moles of O<sub>2</sub> are needed to burn 2 moles of CH<sub>4</sub> in CH<sub>4</sub> + 2O<sub>2</sub> -> CO<sub>2</sub> + 2H<sub>2</sub>O?
Answer: C. 4 mol
Why: The ratio CH<sub>4</sub>:O<sub>2</sub> is 1:2. For 2 moles CH<sub>4</sub>, oxygen needed = 4 moles.
Q63.
What mass of water is formed when 4 g of hydrogen reacts completely with oxygen? Use 2H<sub>2</sub> + O<sub>2</sub> -> 2H<sub>2</sub>O.
Answer: C. 36 g
Why: 4 g H<sub>2</sub> = 2 moles H<sub>2</sub>. The ratio H<sub>2</sub>:H<sub>2</sub>O is 1:1, so 2 moles H<sub>2</sub>O form. Mass = 2 x 18 = 36 g.
Q64.
How many moles of atoms are present in 1 mole of H<sub>2</sub>O molecules?
Answer: C. 3 mol
Why: Each H<sub>2</sub>O molecule has 3 atoms total, so 1 mole of H<sub>2</sub>O contains 3 moles of atoms.
Q65.
Which sample contains the greatest number of molecules?
Answer: D. All contain equal molecules
Why: Each sample is exactly 1 mole of its substance, so each contains 6.022 x 10<sup>23</sup> molecules.
Q66.
How many moles of ions are produced when 1 mole of NaCl dissolves completely in water?
Answer: B. 2 mol
Why: NaCl dissociates into Na+ and Cl-. One mole of NaCl gives 1 mole Na+ and 1 mole Cl-, total 2 moles of ions.
Q67.
What volume of CO<sub>2</sub> at STP is produced from 0.5 mole of CaCO<sub>3</sub> in CaCO<sub>3</sub> -> CaO + CO<sub>2</sub>?
Answer: B. 11.2 L
Why: The ratio CaCO<sub>3</sub>:CO<sub>2</sub> is 1:1, so 0.5 mole CO<sub>2</sub> forms. Volume = 0.5 x 22.4 = 11.2 L.
Q68.
What is the mass of 0.25 mole of H<sub>2</sub>SO<sub>4</sub>?
Answer: B. 24.5 g
Why: Molar mass of H<sub>2</sub>SO<sub>4</sub> is 98 g/mol. Mass = 0.25 x 98 = 24.5 g.
Q69.
How many hydrogen atoms are present in 2 moles of CH<sub>4</sub> molecules?
Answer: C. 8 moles of H atoms
Why: Each CH<sub>4</sub> molecule has 4 hydrogen atoms. 2 moles CH<sub>4</sub> contain 8 moles of hydrogen atoms.
Q70.
A compound has molar mass 60 g/mol. How many molecules are in 30 g of it?
Answer: B. 3.011 x 10<sup>23</sup>
Why: Moles = 30/60 = 0.5 mol. Molecules = 0.5 x 6.022 x 10<sup>23</sup> = 3.011 x 10<sup>23.</sup>
Q71.
A compound contains 80% carbon and 20% hydrogen by mass. Its empirical formula is:
Answer: C. CH₃
Why: C: 80/12 = 6.67; H: 20/1 = 20. Dividing by 6.67 gives C₁H₃, i.e. the empirical formula CH₃.
Q72.
How many grams of NaOH are required to prepare 500 mL of a 0.1 M solution? (NaOH = 40)
Answer: C. 2 g
Why: Moles needed = 0.1 × 0.5 = 0.05; mass = 0.05 × 40 = 2 g.
Q73.
When 4 g of H₂ reacts with 16 g of O₂ (2H₂ + O₂ → 2H₂O), the limiting reagent is:
Answer: B. oxygen
Why: 4 g H₂ = 2 mol and 16 g O₂ = 0.5 mol. The ratio needs 2:1, so 0.5 mol O₂ pairs with only 1 mol H₂ - oxygen runs out first.
Q74.
The vapour density of a gas is 32. Its molar mass is:
Answer: C. 64
Why: Molar mass = 2 × vapour density = 2 × 32 = 64 g/mol.
Q75.
If 0.25 mol of an element weighs 16 g, its atomic mass is:
Answer: C. 64
Why: Atomic mass = mass / moles = 16 / 0.25 = 64 u.