TRUE OR FALSE — THE CONCEPT OF THE MOLE
Relative Atomic Mass & Relative Molecular Mass
1. Relative Atomic Mass (Ar)
Statement 1: Relative atomic mass compares the average mass of an atom of an element to one-twelfth the mass of a carbon-12 atom.
ANSWER: True
Statement 2: Relative atomic mass has a unit, since it is a ratio.
ANSWER: False — Relative atomic mass has no unit; being a ratio, it is dimensionless.
Statement 3: Relative atomic mass accounts for the natural abundance of an element's isotopes.
ANSWER: True
2. Relative Molecular Mass (Mr)
Statement 1: Relative molecular mass is found by summing the relative atomic masses of every atom in a molecule.
ANSWER: True
Statement 2: For ionic compounds, relative molecular mass is instead called relative formula mass.
ANSWER: True
Statement 3: Relative molecular mass is the average mass of one molecule compared with the full mass of a carbon-12 atom.
ANSWER: False — It is compared with one-twelfth the mass of a carbon-12 atom, not the full mass.
3. Atomic Mass Unit (amu)
Statement 1: An atomic mass unit is defined as one-twelfth the mass of a single carbon-12 atom.
ANSWER: True
Statement 2: The atomic mass unit provides the reference scale on which relative atomic mass is built.
ANSWER: True
Statement 3: An atomic mass unit is defined as the full mass of a carbon-12 atom.
ANSWER: False — It is one-twelfth the mass of a carbon-12 atom, not the full mass.
The Mole as a Unit of the Amount of Substance
4. The Mole (mol)
Statement 1: The mole is defined by the number of atoms present in exactly 12 g of carbon-12.
ANSWER: True
Statement 2: The mole is the base SI unit for the amount of substance.
ANSWER: True
Statement 3: The mole represents a variable number of elementary entities that changes from substance to substance.
ANSWER: False — One mole always represents the same fixed number of entities, Avogadro's number, regardless of substance.
5. Elementary Entities
Statement 1: Elementary entities can be atoms, molecules, ions, electrons or protons.
ANSWER: True
Statement 2: Chemists count elementary entities in multiples of Avogadro's number.
ANSWER: True
Statement 3: Elementary entities can only ever refer to atoms, never ions or electrons.
ANSWER: False — The term also covers molecules, ions, electrons and protons.
6. Amount of Substance
Statement 1: Amount of substance is found by dividing the number of entities in a sample by Avogadro's constant.
ANSWER: True
Statement 2: Amount of substance is represented by the symbol n.
ANSWER: True
Statement 3: The base SI unit of amount of substance is the gram.
ANSWER: False — The base SI unit of amount of substance is the mole.
7. Avogadro's Number (Avogadro's Constant)
A sample contains $1.2 \times 10^{24}$ molecules of oxygen. [Avogadro's constant, $L = 6.0 \times 10^{23}\ \text{mol}^{-1}$]
Statement 1: Given the data above, the amount of substance works out to 2.0 mol.
ANSWER: True — $n = \dfrac{N}{L} = \dfrac{1.2 \times 10^{24}}{6.0 \times 10^{23}} = 2.0\ \text{mol}$.
Statement 2: Doubling the number of molecules to $2.4 \times 10^{24}$ would give an amount of substance of 4.0 mol.
ANSWER: True — $n = \dfrac{2.4 \times 10^{24}}{6.0 \times 10^{23}} = 4.0\ \text{mol}$.
Statement 3: Halving the number of molecules to $6.0 \times 10^{23}$ would give an amount of substance of 0.50 mol.
ANSWER: False — $n = \dfrac{6.0 \times 10^{23}}{6.0 \times 10^{23}} = 1.0\ \text{mol}$, not 0.50 mol.
Calculating the Number of Entities, Mass and Volume
8. Number of Entities (N)
A sample contains $0.20\ \text{mol}$ of sodium. [Avogadro's constant, $L = 6.0 \times 10^{23}\ \text{mol}^{-1}$]
Statement 1: Given the data above, the number of sodium atoms works out to $1.2 \times 10^{23}$.
ANSWER: True — $N = n \times L = 0.20 \times (6.0 \times 10^{23}) = 1.2 \times 10^{23}\ \text{atoms}$.
Statement 2: Doubling the amount of substance to 0.40 mol would give $2.4 \times 10^{23}$ entities.
ANSWER: True — $N = 0.40 \times (6.0 \times 10^{23}) = 2.4 \times 10^{23}$.
Statement 3: Using the same 0.20 mol, but with Avogadro's constant instead taken as $3.0 \times 10^{23}\ \text{mol}^{-1}$, would still give $1.2 \times 10^{23}$ entities.
ANSWER: False — $N = 0.20 \times (3.0 \times 10^{23}) = 6.0 \times 10^{22}$, not $1.2 \times 10^{23}$.
9. Molar Mass
A sample of aluminium atoms has a mass of $24\ \text{g}$. [$A_r$ of Al taken as $12$ for this exercise]
Statement 1: Given the data above, the amount of substance of aluminium works out to 2.0 mol.
ANSWER: True — $n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{24}{12} = 2.0\ \text{mol}$.
Statement 2: If the mass were instead 12 g, the amount of substance would be 1.0 mol.
ANSWER: True — $n = \dfrac{12}{12} = 1.0\ \text{mol}$.
Statement 3: If the mass were instead 36 g, the amount of substance would be 4.0 mol.
ANSWER: False — $n = \dfrac{36}{12} = 3.0\ \text{mol}$, not 4.0 mol.
10. Molar Volume (Vm)
A sample contains $1.0\ \text{mol}$ of ammonia gas ($NH_3$) at s.t.p. [$V_m = 22\ \text{dm}^3\text{mol}^{-1}$, taken to 2 s.f.]
Statement 1: Given the data above, the volume of ammonia gas works out to 22 dm³.
ANSWER: True — $V = n \times V_m = 1.0 \times 22 = 22\ \text{dm}^3$.
Statement 2: If the amount of substance were 2.0 mol instead, the volume would be 44 dm³.
ANSWER: True — $V = 2.0 \times 22 = 44\ \text{dm}^3$.
Statement 3: If the amount of substance were 0.50 mol instead, the volume would be 22 dm³.
ANSWER: False — $V = 0.50 \times 22 = 11\ \text{dm}^3$, not 22 dm³.
11. Standard Temperature and Pressure (s.t.p.)
Statement 1: Standard temperature and pressure is fixed at a temperature of 273 K and a pressure of 101.3 kPa.
ANSWER: True
Statement 2: Under s.t.p. conditions, one mole of any gas occupies 22.4 dm³.
ANSWER: True
Statement 3: Standard temperature and pressure varies depending on which gas is being measured.
ANSWER: False — s.t.p. is a fixed reference condition, the same for every gas.
Concentration and Standard Solutions
12. Molarity (Concentration in mol dm⁻³)
A solution has a volume of $500\ \text{cm}^3$ ($0.50\ \text{dm}^3$) and a concentration of $2.0\ \text{mol dm}^{-3}$.
Statement 1: Given the data above, the number of moles present works out to 1.0 mol.
ANSWER: True — $n = C \times V = 2.0 \times 0.50 = 1.0\ \text{mol}$.
Statement 2: If the volume were 250 cm³ (0.25 dm³) instead, the number of moles would be 0.50 mol.
ANSWER: True — $n = 2.0 \times 0.25 = 0.50\ \text{mol}$.
Statement 3: If the concentration were 4.0 mol dm⁻³ instead, over the same 500 cm³, the number of moles would still be 1.0 mol.
ANSWER: False — $n = 4.0 \times 0.50 = 2.0\ \text{mol}$, not 1.0 mol.
13. Mass Concentration
Statement 1: Mass concentration is defined as the mass of solute dissolved in one cubic decimetre of a solution.
ANSWER: True
Statement 2: Mass concentration is expressed in grams per cubic decimetre (g dm⁻³).
ANSWER: True
Statement 3: Mass concentration uses the same symbol as molarity.
ANSWER: False — Mass concentration is represented by a symbol similar to the one used for density, not the symbol used for molarity.
14. Relationship Between Molar and Mass Concentration (C = ρ/M)
Statement 1: The formula C = ρ/M connects molar concentration to mass concentration.
ANSWER: True
Statement 2: In the formula C = ρ/M, M represents molar mass.
ANSWER: True
Statement 3: The formula C = ρ/M cannot be used to switch between grams-per-litre and moles-per-litre thinking.
ANSWER: False — That is exactly what the formula allows chemists to do.
15. Standard Solution
Statement 1: A standard solution's concentration is accurately known.
ANSWER: True
Statement 2: A standard solution can be prepared either from a primary standard or by diluting a concentrated stock solution.
ANSWER: True
Statement 3: A standard solution is unnecessary for accurate volumetric analysis.
ANSWER: False — It is essential for accurate volumetric analysis.
16. Primary Standard
Statement 1: A primary standard is available in pure form or a state of known purity.
ANSWER: True
Statement 2: Sodium carbonate and potassium iodate are examples of a primary standard.
ANSWER: True
Statement 3: A primary standard is expected to react slowly and incompletely.
ANSWER: False — A primary standard should react speedily and completely.
17. Properties of a Primary Standard
Statement 1: A primary standard must be stable and have high solubility.
ANSWER: True
Statement 2: A primary standard must have a reasonably high relative formula mass.
ANSWER: True
Statement 3: A primary standard is allowed to undergo side reactions during standardisation.
ANSWER: False — It must react without side reactions.
18. Preparation of a Standard Solution from a Solid Solute
A learner needs to prepare $500\ \text{cm}^3$ ($0.50\ \text{dm}^3$) of $1.0\ \text{mol dm}^{-3}$ NaOH solution. [$M(\text{NaOH}) = 40\ \text{g mol}^{-1}$]
Statement 1: Given the data above, the number of moles of NaOH required works out to 0.50 mol.
ANSWER: True — $n = C \times V = 1.0 \times 0.50 = 0.50\ \text{mol}$.
Statement 2: Given the data above, the mass of NaOH required works out to 20 g.
ANSWER: True — $m = n \times M = 0.50 \times 40 = 20\ \text{g}$.
Statement 3: If the target concentration were instead 2.0 mol dm⁻³, the mass required would still be 20 g.
ANSWER: False — $n = 2.0 \times 0.50 = 1.0\ \text{mol}$; $m = 1.0 \times 40 = 40\ \text{g}$, not 20 g.
19. Preparation of a Standard Solution from a Concentrated Solution
Statement 1: This preparation starts with a stock solution rather than a solid.
ANSWER: True
Statement 2: This preparation relies on the dilution formula to determine how much stock to measure.
ANSWER: True
Statement 3: This preparation skips the step of topping up to the calibration mark, unlike preparation from a solid.
ANSWER: False — It finishes by topping up to the calibration mark, just like preparation from a solid.
20. The Dilution Formula (C1V1 = C2V2)
A stock solution has a concentration of $12\ \text{mol dm}^{-3}$. A learner wants to prepare $500\ \text{cm}^3$ ($0.50\ \text{dm}^3$) of a $2.0\ \text{mol dm}^{-3}$ solution from it.
Statement 1: Given the data above, the volume of stock solution needed works out to about 83 cm³.
ANSWER: True — $V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{2.0 \times 0.50}{12} = 0.083\ \text{dm}^3 = 83\ \text{cm}^3$.
Statement 2: If the target volume were 250 cm³ (0.25 dm³) instead, the stock volume needed would be about 42 cm³.
ANSWER: True — $V_1 = \dfrac{2.0 \times 0.25}{12} = 0.042\ \text{dm}^3 = 42\ \text{cm}^3$.
Statement 3: If the stock concentration were instead 6.0 mol dm⁻³, the volume needed for the original 500 cm³ target would still be about 83 cm³.
ANSWER: False — $V_1 = \dfrac{2.0 \times 0.50}{6.0} = 0.17\ \text{dm}^3 = 170\ \text{cm}^3$, not 83 cm³.
21. Determination of the Concentration of a Stock Solution
A stock HCl solution has a density of $1.2\ \text{g cm}^{-3}$, a percentage purity of $40\%$, and a molar mass of $40\ \text{g mol}^{-1}$.
Statement 1: Given the data above, the mass of pure HCl in 1 cm³ of stock solution works out to 0.48 g.
ANSWER: True — $1.2 \times 0.40 = 0.48\ \text{g}$.
Statement 2: Given the data above, the concentration of the stock solution works out to 12 mol dm⁻³.
ANSWER: True — $n = \dfrac{0.48}{40} = 0.012\ \text{mol in } 1\ \text{cm}^3$; scaled to $1000\ \text{cm}^3$ gives $12\ \text{mol dm}^{-3}$.
Statement 3: If the percentage purity were instead 20%, the stock concentration would still be 12 mol dm⁻³.
ANSWER: False — mass $= 1.2 \times 0.20 = 0.24\ \text{g}$; $n = 0.24 \div 40 = 0.006\ \text{mol}$; scaled up gives $6.0\ \text{mol dm}^{-3}$, not 12.
22. Percentage Purity
Statement 1: Percentage purity describes how much of a stock solution's mass is actually the substance of interest.
ANSWER: True
Statement 2: Commercial HCl is often only 37% pure.
ANSWER: True
Statement 3: Percentage purity is subtracted from, rather than multiplied by, the total mass of solution to find the mass of pure solute.
ANSWER: False — It is multiplied by the total mass of solution, not subtracted from it.