WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Mathematical Skills in Chemistry
Units, standard form and proportion in chemical calculation: standard form
- 3 things to know
- All units
Overview
Thirteen recall questions on units, standard form and proportion in chemical calculation: standard form.
What you need to know
Answer each one before you reveal it.
Convert 1.25 × 10⁻² mol dm⁻³ to ordinary form without altering the precision implied.
A student converts 2.400 × 10³ g mol⁻¹ to ordinary form and writes 2400 g mol⁻¹. Why is this unsatisfactory?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Write 0.0050 mol dm⁻³ in standard form, and explain how many significant figures the value carries.
Model answer
0.0050 mol dm⁻³ = 5.0 × 10⁻³ mol dm⁻³. The leading zeros only fix the decimal point and are not significant, but the final zero is, so the value has two significant figures. Standard form makes this unambiguous.
- 2
Convert 1.25 × 10⁻² mol dm⁻³ to ordinary form without altering the precision implied.
Model answer
1.25 × 10⁻² mol dm⁻³ = 0.0125 mol dm⁻³. Three significant figures are retained; writing 0.013 would discard information the original value carried.
- 3
A student converts 2.400 × 10³ g mol⁻¹ to ordinary form and writes 2400 g mol⁻¹. Why is this unsatisfactory?
Model answer
The standard form value shows four significant figures, but 2400 written plainly is ambiguous because the trailing zeros could be placeholders. Standard form should be kept, or the precision stated explicitly, so that the reader knows the value is certain to the nearest gram per mole.
- 4
Express the ionic product of water at 298 K in ordinary form and comment on why standard form is preferred.
Model answer
1.00 × 10⁻¹⁴ mol² dm⁻⁶ is 0.0000000000000100 mol² dm⁻⁶ in ordinary form. Standard form is preferred because it is compact, the number of significant figures is unambiguous, and the exponent alone communicates the order of magnitude.
- 5
Calculate the number of molecules in 4.40 g of carbon dioxide. (Avogadro constant = 6.02 × 10²³ mol⁻¹, M(CO₂) = 44.0 g mol⁻¹)
Model answer
n = m ÷ M = 4.40 ÷ 44.0 = 0.100 mol. Number of molecules = 0.100 × 6.02 × 10²³ = 6.02 × 10²² molecules.
- 6
Calculate the mass of a single molecule of water in grams. (M(H₂O) = 18.0 g mol⁻¹, L = 6.02 × 10²³ mol⁻¹)
Model answer
Mass of one molecule = 18.0 ÷ (6.02 × 10²³) = 2.99 × 10⁻²³ g. Dividing by a number in standard form subtracts the exponent, so the answer is very small, as expected for a single molecule.
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