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Stoichiometry and Chemical Calculations MCQs 2026

75 questions with detailed answers · 26 from past papers · 8 quiz batches available

📚 Chemistry Mcqs 📄 26 Past-Paper Qs ✓ Free · No Login Needed
🎯 Mock Test

Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1 Questions 1–10 of 75
  1. Q1 easy

    One mole of any substance contains ___ particles

    1. A 6.022×10^22
    2. B 3.011×10^23
    3. C 1×10^23
    4. D 6.022×10^23
    💡 Explanation:

    Avogadro's number, 6.022×10^23, defines the number of particles in one mole.

  2. Q2 Past Paper · PPSC/FPSC/NTS easy

    The molar mass of water (H2O) is

    1. A 16 g/mol
    2. B 20 g/mol
    3. C 18 g/mol
    4. D 17 g/mol
    💡 Explanation:

    Water's molar mass is (2×1) + 16 = 18 g/mol.

  3. Q3 Past Paper · PPSC/FPSC/NTS easy

    The number of moles in 44 g of CO2 (molar mass 44 g/mol) is

    1. A 0.5 mol
    2. B 2 mol
    3. C 4 mol
    4. D 1 mol
    💡 Explanation:

    Moles = mass ÷ molar mass = 44/44 = 1 mole.

  4. Q4 easy

    The mass of 2 moles of NaOH (molar mass 40 g/mol) is

    1. A 80 g
    2. B 40 g
    3. C 20 g
    4. D 100 g
    💡 Explanation:

    Mass = moles × molar mass = 2 × 40 = 80 g.

  5. Q5 Past Paper · PPSC/FPSC/NTS easy

    The number of atoms in 1 mole of helium gas is

    1. A 3.011×10^23
    2. B 12.044×10^23
    3. C 1×10^23
    4. D 6.022×10^23
    💡 Explanation:

    One mole of any element contains Avogadro's number of atoms, 6.022×10^23.

  6. Q6 easy

    Avogadro's number is approximately

    1. A 3.14×10^23
    2. B 6.022×10^23
    3. C 9.8×10^23
    4. D 1.6×10^-19
    💡 Explanation:

    Avogadro's number is defined as 6.022×10^23 particles per mole.

  7. Q7 Past Paper · PPSC/FPSC/NTS easy

    The volume occupied by 1 mole of an ideal gas at STP is

    1. A 24 L
    2. B 25 L
    3. C 20 L
    4. D 22.4 L
    💡 Explanation:

    At standard temperature and pressure, one mole of ideal gas occupies 22.4 liters.

  8. Q8 easy

    The empirical formula represents

    1. A The simplest whole-number ratio of atoms in a compound
    2. B The exact number of atoms in one molecule
    3. C The molar mass of the compound
    4. D The number of moles present
    💡 Explanation:

    Empirical formulas give the simplest ratio of elements, not the actual molecular composition.

  9. Q9 Past Paper · PPSC/FPSC/NTS easy

    The molecular formula of a compound is always

    1. A A whole-number multiple of the empirical formula
    2. B Smaller than the empirical formula
    3. C Unrelated to the empirical formula
    4. D Equal to half the empirical formula
    💡 Explanation:

    Molecular formula equals the empirical formula multiplied by a whole number, including 1.

  10. Q10 easy

    If the empirical formula of a compound is CH2O and its molar mass is 180 g/mol, its molecular formula is

    1. A C2H4O2
    2. B C3H6O3
    3. C C6H12O6
    4. D CH2O
    💡 Explanation:

    Empirical formula mass is 30; 180÷30=6, so the molecular formula is (CH2O)×6 = C6H12O6.

  11. Q11 medium

    The percentage composition of an element in a compound is calculated as

    1. A (Mass of element in one mole ÷ Molar mass of compound) × 100
    2. B (Moles of compound ÷ Avogadro's number) × 100
    3. C Atomic mass ÷ 100
    4. D Molar mass ÷ mass of element
    💡 Explanation:

    Percent composition compares the mass contributed by an element to the total molar mass.

  12. Q12 Past Paper · PPSC/FPSC/NTS medium

    The percentage of oxygen by mass in water (H2O, molar mass 18) is approximately

    1. A 88.9%
    2. B 11.1%
    3. C 50%
    4. D 66.7%
    💡 Explanation:

    Oxygen contributes 16 out of 18 g/mol, giving 16/18 × 100 ≈ 88.9%.

  13. Q13 medium

    The limiting reagent in a reaction is the reactant that

    1. A Is present in excess
    2. B Does not react
    3. C Is completely consumed first, limiting the amount of product
    4. D Has the highest molar mass
    💡 Explanation:

    The reaction stops producing more product once the limiting reagent is used up.

  14. Q14 Past Paper · PPSC/FPSC/NTS medium

    The reactant that remains after a reaction is complete is called the

    1. A Excess reagent
    2. B Limiting reagent
    3. C Catalyst
    4. D Product
    💡 Explanation:

    Any reactant left over after the limiting reagent is consumed is called the excess reagent.

  15. Q15 medium

    Theoretical yield refers to

    1. A The actual amount of product obtained in the lab
    2. B The percentage of reactant used
    3. C The maximum amount of product predicted by stoichiometric calculation
    4. D The amount of limiting reagent
    💡 Explanation:

    Theoretical yield is the calculated maximum product possible assuming complete reaction.

  16. Q16 medium

    Percentage yield is calculated as

    1. A Theoretical yield ÷ molar mass × 100
    2. B (Actual yield ÷ Theoretical yield) × 100
    3. C (Theoretical yield ÷ Actual yield) × 100
    4. D Actual yield × molar mass
    💡 Explanation:

    Percentage yield compares the actual product obtained to the theoretically possible amount.

  17. Q17 Past Paper · PPSC/FPSC/NTS medium

    A balanced chemical equation obeys the law of

    1. A Definite proportions only
    2. B Multiple proportions only
    3. C Avogadro's law
    4. D Conservation of mass
    💡 Explanation:

    Balancing ensures equal numbers of atoms on both sides, honoring conservation of mass.

  18. Q18 medium

    In the reaction N2 + 3H2 → 2NH3, the mole ratio of H2 to NH3 is

    1. A 1:1
    2. B 1:2
    3. C 2:3
    4. D 3:2
    💡 Explanation:

    The balanced equation shows 3 moles of H2 react to form 2 moles of NH3, a 3:2 ratio.

  19. Q19 Past Paper · PPSC/FPSC/NTS hard

    The formula weight of MgO (Mg=24, O=16) is

    1. A 32
    2. B 40
    3. C 56
    4. D 48
    💡 Explanation:

    Sum of atomic masses: 24+16 = 40.

  20. Q20 medium

    The number of moles of oxygen atoms in 2 moles of CO2 is

    1. A 4 mol
    2. B 2 mol
    3. C 1 mol
    4. D 6 mol
    💡 Explanation:

    Each CO2 molecule has 2 oxygen atoms, so 2 moles of CO2 contain 4 moles of oxygen atoms.

  21. Q21 Past Paper · PPSC/FPSC/NTS medium

    Molarity of a solution is defined as

    1. A Moles of solute per kg of solvent
    2. B Moles of solute per liter of solution
    3. C Grams of solute per liter of solution
    4. D Moles of solvent per liter of solution
    💡 Explanation:

    Molarity (M) is defined as moles of solute divided by liters of total solution.

  22. Q22 medium

    The molarity of a solution containing 2 moles of NaCl in 500 mL of solution is

    1. A 1 M
    2. B 0.5 M
    3. C 2 M
    4. D 4 M
    💡 Explanation:

    Molarity = 2 mol ÷ 0.5 L = 4 M.

  23. Q23 medium

    Molality of a solution is defined as

    1. A Moles of solute per liter of solution
    2. B Grams of solute per liter of solvent
    3. C Moles of solvent per kg of solute
    4. D Moles of solute per kilogram of solvent
    💡 Explanation:

    Molality (m) is moles of solute per kilogram of solvent, independent of temperature-based volume changes.

  24. Q24 Past Paper · PPSC/FPSC/NTS medium

    The number of moles of solute in 250 mL of a 2 M solution is

    1. A 1 mol
    2. B 0.5 mol
    3. C 2 mol
    4. D 0.25 mol
    💡 Explanation:

    Moles = molarity × volume(L) = 2 × 0.25 = 0.5 mol.

  25. Q25 medium

    The mass of 0.25 mol of CaCO3 (molar mass 100 g/mol) is

    1. A 100 g
    2. B 4 g
    3. C 25 g
    4. D 50 g
    💡 Explanation:

    Mass = moles × molar mass = 0.25 × 100 = 25 g.

  26. Q26 medium

    The molar mass of glucose, C6H12O6, is

    1. A 180 g/mol
    2. B 162 g/mol
    3. C 194 g/mol
    4. D 176 g/mol
    💡 Explanation:

    Summing atomic masses: (6×12)+(12×1)+(6×16) = 72+12+96 = 180 g/mol.

  27. Q27 Past Paper · PPSC/FPSC/NTS medium

    The number of moles present in 11.2 L of a gas at STP is

    1. A 1 mol
    2. B 2 mol
    3. C 0.25 mol
    4. D 0.5 mol
    💡 Explanation:

    Moles = volume ÷ 22.4 L/mol = 11.2/22.4 = 0.5 mol.

  28. Q28 medium

    According to the law of conservation of mass, in a chemical reaction the total mass of reactants

    1. A Increases
    2. B Equals the total mass of products
    3. C Decreases
    4. D Becomes zero
    💡 Explanation:

    Mass is neither created nor destroyed, so total reactant mass equals total product mass.

  29. Q29 medium

    The law of definite proportions states that a chemical compound always contains the same elements

    1. A In variable proportions by mass
    2. B In fixed proportions by mass
    3. C In equal number of moles only
    4. D Only in the gaseous state
    💡 Explanation:

    A pure compound always has the same elements combined in the same fixed mass ratio.

  30. Q30 Past Paper · PPSC/FPSC/NTS medium

    The law of multiple proportions applies when two elements form

    1. A Only one compound
    2. B Ionic compounds only
    3. C More than one compound, with masses in small whole-number ratios
    4. D Isotopes
    💡 Explanation:

    When two elements form multiple compounds, the masses of one element combining with a fixed mass of the other form small whole-number ratios.

  31. Q31 medium

    The number of moles of NaOH needed to completely neutralize 1 mole of H2SO4 is

    1. A 1 mol
    2. B 0.5 mol
    3. C 2 mol
    4. D 3 mol
    💡 Explanation:

    H2SO4 is diprotic, requiring 2 moles of NaOH for complete neutralization.

  32. Q32 medium

    In the reaction 2H2 + O2 → 2H2O, the number of moles of water produced from 4 moles of H2 (excess O2) is

    1. A 2 mol
    2. B 1 mol
    3. C 4 mol
    4. D 8 mol
    💡 Explanation:

    The 2:2 mole ratio of H2 to H2O means 4 moles of H2 produce 4 moles of water.

  33. Q33 Past Paper · PPSC/FPSC/NTS medium

    A 1 molar solution of a substance contains how many moles of solute per liter

    1. A 0.5 mol
    2. B 1 mol
    3. C 2 mol
    4. D 10 mol
    💡 Explanation:

    By definition, a 1 M solution contains exactly 1 mole of solute per liter of solution.

  34. Q34 medium

    Which quantity remains constant during a chemical reaction, per the law of conservation of mass

    1. A Total mass
    2. B Volume of gas only
    3. C Number of moles of gas
    4. D Temperature
    💡 Explanation:

    Total mass of the reacting system stays constant throughout a chemical reaction.

  35. Q35 medium

    The formula mass of Na2CO3 (Na=23, C=12, O=16) is

    1. A 84
    2. B 106
    3. C 90
    4. D 62
    💡 Explanation:

    Sum: (2×23)+12+(3×16) = 46+12+48 = 106.

  36. Q36 Past Paper · PPSC/FPSC/NTS medium

    The percentage of carbon by mass in CO2 (molar mass 44) is approximately

    1. A 50%
    2. B 12%
    3. C 27.3%
    4. D 73%
    💡 Explanation:

    Carbon contributes 12 out of 44 g/mol, giving 12/44 × 100 ≈ 27.3%.

  37. Q37 medium

    In stoichiometric calculations, mole ratios are obtained directly from the

    1. A Atomic masses
    2. B Coefficients of the balanced equation
    3. C Molar volumes
    4. D Percentage composition
    💡 Explanation:

    The coefficients in a balanced equation give the exact mole ratio between reactants and products.

  38. Q38 medium

    If 5 moles of a limiting reagent produce 5 moles of product in a 1:1 reaction and only 4 moles of product were actually obtained, the percentage yield is

    1. A 100%
    2. B 125%
    3. C 80%
    4. D 50%
    💡 Explanation:

    Percentage yield = (4/5) × 100 = 80%.

  39. Q39 Past Paper · PPSC/FPSC/NTS medium

    A solution containing 1 mole of solute dissolved in exactly 1 kg of solvent has a molality of

    1. A 0.5 m
    2. B 1 m
    3. C 2 m
    4. D 10 m
    💡 Explanation:

    By definition, molality equals moles of solute per kilogram of solvent, giving 1 m here.

  40. Q40 medium

    The relative formula mass of a compound is the sum of the

    1. A Atomic numbers of all atoms
    2. B Number of moles of each element
    3. C Charges of all ions
    4. D Relative atomic masses of all atoms in the formula
    💡 Explanation:

    Formula mass is calculated by adding up the atomic masses of every atom shown in the formula.

  41. Q41 medium

    Which of the following best describes an empirical formula determination experiment

    1. A Measuring the boiling point of a compound
    2. B Measuring the volume of gas released
    3. C Titrating an acid with a base
    4. D Finding the mass ratio of elements combined in a compound
    💡 Explanation:

    Empirical formula experiments typically measure the mass of each element present to find their simplest ratio.

  42. Q42 Past Paper · PPSC/FPSC/NTS hard

    The number of moles of ions produced when 1 mole of Na2SO4 fully dissociates in water is

    1. A 1 mol
    2. B 2 mol
    3. C 4 mol
    4. D 3 mol
    💡 Explanation:

    Na2SO4 dissociates into 2 Na+ ions and 1 SO4^2- ion, totaling 3 moles of ions.

  43. Q43 hard

    The mass percent of hydrogen in methane, CH4 (molar mass 16 g/mol) is approximately

    1. A 12.5%
    2. B 20%
    3. C 25%
    4. D 75%
    💡 Explanation:

    Hydrogen contributes 4 out of 16 g/mol, giving 4/16 × 100 = 25%.

  44. Q44 hard

    1 mole of a diatomic gas at STP occupies a volume of

    1. A 22.4 L
    2. B 11.2 L
    3. C 44.8 L
    4. D 6.022 L
    💡 Explanation:

    Molar volume at STP is 22.4 L for any ideal gas, regardless of whether it is diatomic.

  45. Q45 hard

    How many moles of oxygen gas (O2) are required to completely combust 1 mole of methane (CH4 + 2O2 → CO2 + 2H2O)

    1. A 1 mol
    2. B 2 mol
    3. C 3 mol
    4. D 0.5 mol
    💡 Explanation:

    The balanced equation shows 2 moles of O2 are needed per mole of methane burned.

  46. Q46 hard

    The number of grams in 0.5 mole of oxygen gas, O2 (molar mass 32 g/mol) is

    1. A 32 g
    2. B 16 g
    3. C 8 g
    4. D 64 g
    💡 Explanation:

    Mass = moles × molar mass = 0.5 × 32 = 16 g.

  47. Q47 Past Paper · PPSC/FPSC/NTS hard

    Gram atomic mass is the mass, in grams, of

    1. A One mole of atoms of an element
    2. B One atom of an element
    3. C One molecule of a compound
    4. D One ion of an element
    💡 Explanation:

    Gram atomic mass equals the atomic mass expressed in grams for one mole of atoms.

  48. Q48 hard

    The number of moles of electrons required to reduce 1 mole of Al3+ to Al metal is

    1. A 3 mol
    2. B 1 mol
    3. C 2 mol
    4. D 6 mol
    💡 Explanation:

    Al3+ requires gaining 3 electrons per ion to become neutral aluminum metal.

  49. Q49 hard

    Stoichiometry is fundamentally based on the relationship between

    1. A Temperature and pressure of gases only
    2. B Enthalpy and entropy
    3. C Atomic radius and ionization energy
    4. D Moles of reactants and products in a balanced equation
    💡 Explanation:

    Stoichiometric calculations use the mole ratios from a balanced chemical equation.

  50. Q50 hard

    The number of moles in 9 g of water (molar mass 18 g/mol) is

    1. A 1 mol
    2. B 0.25 mol
    3. C 0.5 mol
    4. D 2 mol
    💡 Explanation:

    Moles = mass ÷ molar mass = 9/18 = 0.5 mol.

  51. Q51 medium

    The number of moles of oxygen atoms in 2 moles of CO2 is

    1. A 4 mol
    2. B 2 mol
    3. C 1 mol
    4. D 6 mol
    💡 Explanation:

    Each CO2 molecule has 2 oxygen atoms, so 2 moles of CO2 contain 4 moles of oxygen atoms.

  52. Q52 Past Paper · PPSC/FPSC/NTS medium

    Molarity of a solution is defined as

    1. A Moles of solute per kg of solvent
    2. B Moles of solute per liter of solution
    3. C Grams of solute per liter of solution
    4. D Moles of solvent per liter of solution
    💡 Explanation:

    Molarity (M) is defined as moles of solute divided by liters of total solution.

  53. Q53 medium

    The molarity of a solution containing 2 moles of NaCl in 500 mL of solution is

    1. A 1 M
    2. B 0.5 M
    3. C 2 M
    4. D 4 M
    💡 Explanation:

    Molarity = 2 mol ÷ 0.5 L = 4 M.

  54. Q54 medium

    Molality of a solution is defined as

    1. A Moles of solute per liter of solution
    2. B Grams of solute per liter of solvent
    3. C Moles of solvent per kg of solute
    4. D Moles of solute per kilogram of solvent
    💡 Explanation:

    Molality (m) is moles of solute per kilogram of solvent, independent of temperature-based volume changes.

  55. Q55 Past Paper · PPSC/FPSC/NTS medium

    The number of moles of solute in 250 mL of a 2 M solution is

    1. A 1 mol
    2. B 0.5 mol
    3. C 2 mol
    4. D 0.25 mol
    💡 Explanation:

    Moles = molarity × volume(L) = 2 × 0.25 = 0.5 mol.

  56. Q56 medium

    The mass of 0.25 mol of CaCO3 (molar mass 100 g/mol) is

    1. A 100 g
    2. B 4 g
    3. C 25 g
    4. D 50 g
    💡 Explanation:

    Mass = moles × molar mass = 0.25 × 100 = 25 g.

  57. Q57 medium

    The molar mass of glucose, C6H12O6, is

    1. A 180 g/mol
    2. B 162 g/mol
    3. C 194 g/mol
    4. D 176 g/mol
    💡 Explanation:

    Summing atomic masses: (6×12)+(12×1)+(6×16) = 72+12+96 = 180 g/mol.

  58. Q58 Past Paper · PPSC/FPSC/NTS medium

    The number of moles present in 11.2 L of a gas at STP is

    1. A 1 mol
    2. B 2 mol
    3. C 0.25 mol
    4. D 0.5 mol
    💡 Explanation:

    Moles = volume ÷ 22.4 L/mol = 11.2/22.4 = 0.5 mol.

  59. Q59 medium

    According to the law of conservation of mass, in a chemical reaction the total mass of reactants

    1. A Increases
    2. B Equals the total mass of products
    3. C Decreases
    4. D Becomes zero
    💡 Explanation:

    Mass is neither created nor destroyed, so total reactant mass equals total product mass.

  60. Q60 medium

    The law of definite proportions states that a chemical compound always contains the same elements

    1. A In variable proportions by mass
    2. B In fixed proportions by mass
    3. C In equal number of moles only
    4. D Only in the gaseous state
    💡 Explanation:

    A pure compound always has the same elements combined in the same fixed mass ratio.

  61. Q61 Past Paper · PPSC/FPSC/NTS medium

    The law of multiple proportions applies when two elements form

    1. A Only one compound
    2. B Ionic compounds only
    3. C More than one compound, with masses in small whole-number ratios
    4. D Isotopes
    💡 Explanation:

    When two elements form multiple compounds, the masses of one element combining with a fixed mass of the other form small whole-number ratios.

  62. Q62 medium

    The number of moles of NaOH needed to completely neutralize 1 mole of H2SO4 is

    1. A 1 mol
    2. B 0.5 mol
    3. C 2 mol
    4. D 3 mol
    💡 Explanation:

    H2SO4 is diprotic, requiring 2 moles of NaOH for complete neutralization.

  63. Q63 medium

    In the reaction 2H2 + O2 → 2H2O, the number of moles of water produced from 4 moles of H2 (excess O2) is

    1. A 2 mol
    2. B 1 mol
    3. C 4 mol
    4. D 8 mol
    💡 Explanation:

    The 2:2 mole ratio of H2 to H2O means 4 moles of H2 produce 4 moles of water.

  64. Q64 Past Paper · PPSC/FPSC/NTS medium

    A 1 molar solution of a substance contains how many moles of solute per liter

    1. A 0.5 mol
    2. B 1 mol
    3. C 2 mol
    4. D 10 mol
    💡 Explanation:

    By definition, a 1 M solution contains exactly 1 mole of solute per liter of solution.

  65. Q65 medium

    Which quantity remains constant during a chemical reaction, per the law of conservation of mass

    1. A Total mass
    2. B Volume of gas only
    3. C Number of moles of gas
    4. D Temperature
    💡 Explanation:

    Total mass of the reacting system stays constant throughout a chemical reaction.

  66. Q66 medium

    The formula mass of Na2CO3 (Na=23, C=12, O=16) is

    1. A 84
    2. B 106
    3. C 90
    4. D 62
    💡 Explanation:

    Sum: (2×23)+12+(3×16) = 46+12+48 = 106.

  67. Q67 Past Paper · PPSC/FPSC/NTS medium

    The percentage of carbon by mass in CO2 (molar mass 44) is approximately

    1. A 50%
    2. B 12%
    3. C 27.3%
    4. D 73%
    💡 Explanation:

    Carbon contributes 12 out of 44 g/mol, giving 12/44 × 100 ≈ 27.3%.

  68. Q68 medium

    In stoichiometric calculations, mole ratios are obtained directly from the

    1. A Atomic masses
    2. B Coefficients of the balanced equation
    3. C Molar volumes
    4. D Percentage composition
    💡 Explanation:

    The coefficients in a balanced equation give the exact mole ratio between reactants and products.

  69. Q69 medium

    If 5 moles of a limiting reagent produce 5 moles of product in a 1:1 reaction and only 4 moles of product were actually obtained, the percentage yield is

    1. A 100%
    2. B 125%
    3. C 80%
    4. D 50%
    💡 Explanation:

    Percentage yield = (4/5) × 100 = 80%.

  70. Q70 Past Paper · PPSC/FPSC/NTS medium

    A solution containing 1 mole of solute dissolved in exactly 1 kg of solvent has a molality of

    1. A 0.5 m
    2. B 1 m
    3. C 2 m
    4. D 10 m
    💡 Explanation:

    By definition, molality equals moles of solute per kilogram of solvent, giving 1 m here.

  71. Q71 medium

    The relative formula mass of a compound is the sum of the

    1. A Atomic numbers of all atoms
    2. B Number of moles of each element
    3. C Charges of all ions
    4. D Relative atomic masses of all atoms in the formula
    💡 Explanation:

    Formula mass is calculated by adding up the atomic masses of every atom shown in the formula.

  72. Q72 medium

    Which of the following best describes an empirical formula determination experiment

    1. A Measuring the boiling point of a compound
    2. B Measuring the volume of gas released
    3. C Titrating an acid with a base
    4. D Finding the mass ratio of elements combined in a compound
    💡 Explanation:

    Empirical formula experiments typically measure the mass of each element present to find their simplest ratio.

  73. Q73 Past Paper · PPSC/FPSC/NTS hard

    The number of moles of ions produced when 1 mole of Na2SO4 fully dissociates in water is

    1. A 1 mol
    2. B 2 mol
    3. C 4 mol
    4. D 3 mol
    💡 Explanation:

    Na2SO4 dissociates into 2 Na+ ions and 1 SO4^2- ion, totaling 3 moles of ions.

  74. Q74 hard

    The mass percent of hydrogen in methane, CH4 (molar mass 16 g/mol) is approximately

    1. A 12.5%
    2. B 20%
    3. C 25%
    4. D 75%
    💡 Explanation:

    Hydrogen contributes 4 out of 16 g/mol, giving 4/16 × 100 = 25%.

  75. Q75 hard

    1 mole of a diatomic gas at STP occupies a volume of

    1. A 22.4 L
    2. B 11.2 L
    3. C 44.8 L
    4. D 6.022 L
    💡 Explanation:

    Molar volume at STP is 22.4 L for any ideal gas, regardless of whether it is diatomic.

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