CC01 — Atoms, Molecules & Chemical Arithmetic

📗 CDS General Knowledge20 Questions · No Negative Marking

Question 1 of 20

Which law states that in a chemical reaction, total mass of reactants equals total mass of products?

ALaw of Definite Proportions BLaw of Conservation of Mass (Lavoisier) CLaw of Multiple Proportions DAvogadro's Law

Law of Conservation of Mass (Lavoisier, 1789): matter can neither be created nor destroyed in a chemical reaction — only rearranged. This is why we balance chemical equations. The total mass of reactants = total mass of products, even though atoms are rearranged.

Question 2 of 20

One mole of any substance contains how many particles?

A3.011 × 10²³ B6.022 × 10²³ (Avogadro's number) C1.6 × 10⁻¹⁹ D9.1 × 10⁻³¹

Avogadro's number (Nₐ) = 6.022 × 10²³ mol⁻¹. One mole of any substance contains this many particles (atoms, molecules, or ions). Example: 1 mole of H₂O = 6.022 × 10²³ molecules. Established so that atomic masses in g/mol are numerically equal to their relative atomic masses.

Question 3 of 20

The molar mass of H₂SO₄ (sulphuric acid) is:

A80 g/mol B96 g/mol C98 g/mol D100 g/mol

H₂SO₄: (2×1) + (1×32) + (4×16) = 2 + 32 + 64 = 98 g/mol. Atomic masses: H=1, S=32, O=16. Sulphuric acid (98 g/mol) is the most important industrial chemical. 1 mole of H₂SO₄ = 98 g = 6.022 × 10²³ molecules.

Question 4 of 20

At STP, the volume occupied by 1 mole of any ideal gas is:

A11.2 L B22.4 L C44.8 L D24.8 L

At STP (0°C, 1 atm): 1 mole of any ideal gas = 22.4 L (molar volume). This follows from the ideal gas law: PV = nRT → V = nRT/P = (1)(0.0821)(273)/1 = 22.4 L. This is the most frequently tested gas law value in CDS Chemistry.

Question 5 of 20

The empirical formula CH₂O could represent which of the following molecules?

AFormaldehyde (CH₂O, MW=30) only BGlucose, acetic acid, or formaldehyde CMethanol (CH₃OH) only DEthanol (C₂H₅OH) only

The empirical formula CH₂O (C:H:O = 1:2:1) is shared by: formaldehyde (n=1, MW=30), acetic acid (n=2, MW=60), lactic acid (n=3, MW=90), glucose (n=6, MW=180). The molecular formula = n × empirical formula, where n is determined from the molar mass.

Question 6 of 20

How many moles are present in 36 g of water (H₂O, molar mass = 18 g/mol)?

A1 mole B2 moles C3 moles D0.5 moles

Moles = mass / molar mass = 36 / 18 = 2 moles. 2 moles of H₂O = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules. This is a standard stoichiometry calculation — practice mole calculations as they appear frequently in CDS.

Question 7 of 20

The law of definite proportions states that a pure compound always contains its elements:

AIn whole number ratios that differ between samples BIn fixed proportions by mass, regardless of source or method of preparation CIn proportions that change with temperature DIn ratios determined by the atomic number of each element

Law of Definite Proportions (Proust, 1799): a pure compound always has the same elements in the same fixed mass ratio. Example: water always has H:O by mass = 1:8, whether from rain, tap, or synthesis in a lab. This is why we can write a definite chemical formula for each compound.

Question 8 of 20

The percentage of oxygen in water (H₂O) by mass is:

A11.1% B50% C66.7% D88.9%

% O in H₂O = (16/18) × 100 = 88.9%. % H = (2/18) × 100 = 11.1%. Total = 100%. Check: 88.9 + 11.1 = 100%. This is a standard percentage composition calculation. Molar mass of H₂O = 2(1) + 16 = 18 g/mol.

Question 9 of 20

Molarity of a solution is defined as:

AGrams of solute per 100 mL of solution BMoles of solute per kilogram of solvent (molality) CMoles of solute per litre of solution DMass fraction of solute in solution

Molarity (M) = moles of solute ÷ volume of solution in litres. Unit: mol/L or M. Example: 2 M NaCl = 2 moles of NaCl dissolved in enough water to make 1 litre of solution. Note: molality = moles per kg of solvent (not volume, not solution).

Question 10 of 20

In the balanced equation: 2H₂ + O₂ → 2H₂O, the molar ratio of H₂ to O₂ is:

A1:1 B2:1 C1:2 D3:1

Coefficients in a balanced equation give the molar ratios: 2 moles H₂ : 1 mole O₂ : 2 moles H₂O. So H₂:O₂ = 2:1. If 4 moles of H₂ are burned, 2 moles of O₂ are needed and 4 moles of H₂O are produced. Stoichiometric ratios from balanced equations are fundamental to quantitative chemistry.

Question 11 of 20

The mass of 3 moles of CO₂ (molar mass = 44 g/mol) is:

A44 g B88 g C132 g D176 g

Mass = moles × molar mass = 3 × 44 = 132 g. CO₂: C=12, O=16 (×2=32), total = 44 g/mol. This is the basic mole-mass conversion. Always: mass (g) = n (mol) × M (g/mol).

Question 12 of 20

Avogadro's Law states that at the same temperature and pressure, equal volumes of gases:

AHave equal masses BContain equal numbers of molecules CHave equal densities DHave equal molar masses

Avogadro's Law (1811): equal volumes of all gases at the same T and P contain equal numbers of molecules. This explains molar volume (22.4 L = 1 mole at STP). It also shows that 1 L of H₂ and 1 L of CO₂ at the same T and P contain the same number of molecules, even though they have very different masses.

Question 13 of 20

If the formula of a compound is MgCl₂, which of the following is the correct ionic charge of Mg?

AMg²⁻ BMg⁺ CMg²⁺ DMg³⁺

In MgCl₂: Cl has charge −1 (×2 = −2 total). For the compound to be electrically neutral: Mg must be +2. So Mg = Mg²⁺. This is consistent with Mg being in Group 2 of the periodic table — it loses 2 electrons to achieve the stable noble gas configuration of neon.

Question 14 of 20

The number of atoms in 32 g of sulphur (atomic mass = 32 u) is:

A6.022 × 10²³ B3.011 × 10²³ C1.204 × 10²⁴ D1.6 × 10²³

32 g of S = 32/32 = 1 mole of S atoms = 6.022 × 10²³ atoms. Moles = mass/atomic mass = 32/32 = 1 mol. Number of atoms = 1 × Nₐ = 6.022 × 10²³. This is a direct application of the mole concept.

Question 15 of 20

Dalton's atomic theory stated that atoms of different elements:

AHave identical masses BCombine in variable ratios to form compounds CHave different masses and properties DCan be subdivided into protons and electrons

Dalton's Atomic Theory (1808): (1) all matter is made of indivisible atoms, (2) atoms of same element are identical in mass and properties, (3) atoms of different elements differ in mass and properties, (4) atoms combine in simple whole-number ratios, (5) atoms cannot be created or destroyed. Exceptions discovered later (isotopes, nuclear reactions).

Question 16 of 20

The molecular formula of glucose is C₆H₁₂O₆. How many hydrogen atoms are in 1 mole of glucose?

A6.022 × 10²³ B1.2044 × 10²⁴ C7.226 × 10²⁴ D3.613 × 10²⁴

1 molecule of glucose has 12 H atoms. 1 mole of glucose = 6.022 × 10²³ molecules. H atoms = 12 × 6.022 × 10²³ = 7.226 × 10²⁴ atoms. For any molecule CₓHyOz: atoms of H in 1 mole = y × Nₐ.

Question 17 of 20

The mass of one atom of carbon-12 (ⁱ²C) is exactly:

A1 atomic mass unit (u) B12 atomic mass units (u) C6 atomic mass units (u) D24 atomic mass units (u)

By definition, 1 atomic mass unit (u or amu) = 1/12 the mass of one carbon-12 atom. Therefore, one atom of C-12 has a mass of exactly 12 u. This is the international standard for defining the atomic mass unit. 1 u = 1.66 × 10⁻²⁷ kg.

Question 18 of 20

In the reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, if 10 g of CaCO₃ (molar mass = 100 g/mol) reacts completely, the volume of CO₂ at STP is:

A1.12 L B2.24 L C4.48 L D22.4 L

Moles of CaCO₃ = 10/100 = 0.1 mol. From equation: 1 mol CaCO₃ → 1 mol CO₂. So 0.1 mol CO₂ produced. Volume at STP = 0.1 × 22.4 = 2.24 L. This is a complete stoichiometry calculation — moles from mass → moles of product → volume at STP.

Question 19 of 20

The gram-equivalent weight concept states that one equivalent of any substance:

AIs exactly 1 gram regardless of which substance is involved BReacts with or displaces 1 gram of hydrogen (or 8 g of oxygen) CContains exactly 6.022 × 10²³ particles of that substance DIs always the molar mass of the substance divided by exactly 2

Equivalent weight = molar mass / valency (or number of H replaced for acids, or OH replaced for bases). Example: H₂SO₄ equivalent weight = 98/2 = 49 g/eq. One equivalent of acid neutralises one equivalent of base. Used in titration calculations: N₁V₁ = N₂V₂ (normality equation).

Question 20 of 20

Which of the following has the largest number of molecules?

A18 g of H₂O B28 g of CO C16 g of CH₄ D32 g of SO₂

Calculate moles: H₂O: 18/18 = 1 mol; CO: 28/28 = 1 mol; CH₄: 16/16 = 1 mol; SO₂: 32/64 = 0.5 mol. H₂O, CO, and CH₄ all have 1 mole = equal molecules (6.022 × 10²³). SO₂ has only 0.5 mol. All three (A, B, C) are equal — but the question tests if students notice that SO₂ has fewer. Trick: H₂O, CO, and CH₄ are tied; SO₂ is less.