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: volume, equilibrium, rate and energy units
- 3 things to know
- All units
Overview
Converting volumes for c = n/V and pV = nRT, deducing the units of Kc and of rate constants, matching kJ and J in Gibbs calculations, and the units of Kw and Ka.
What you need to know
Answer each one before you reveal it.
A student writes that 250 cm³ of solution is 0.250 m³. What has gone wrong, and what are the correct values in dm³ and m³?
Why must volume be in m³ rather than dm³ when the ideal gas equation is used with pressure in pascals?
Questions from this subtopic
Each one with the model answer underneath.
- 1
A pipette delivers 25.0 cm³ of a solution. What is this volume in dm³, and why does the conversion matter before using c = n/V?
Model answer
25.0 cm³ is 25.0 ÷ 1000 = 0.0250 dm³, or 2.50 × 10⁻² dm³. The conversion matters because concentration in chemistry is quoted in mol dm⁻³, so the volume substituted into n = c × V must be in dm³; leaving it in cm³ inflates the number of moles by a factor of 1000.
- 2
Calculate the amount, in moles, of solute in 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.
Model answer
V = 25.0 cm³ = 0.0250 dm³. n = c × V = 0.100 mol dm⁻³ × 0.0250 dm³ = 2.50 × 10⁻³ mol. The units cancel to leave mol, which confirms the conversion was made correctly.
- 3
A student writes that 250 cm³ of solution is 0.250 m³. What has gone wrong, and what are the correct values in dm³ and m³?
Model answer
The student has used the cm³ to dm³ factor but written the answer as m³. 250 cm³ = 0.250 dm³, and since 1 dm³ = 1 × 10⁻³ m³, the volume is 2.50 × 10⁻⁴ m³. Three different volume units are in routine use and each conversion step must be tracked separately.
- 4
Why must volume be in m³ rather than dm³ when the ideal gas equation is used with pressure in pascals?
Model answer
R = 8.31 J K⁻¹ mol⁻¹ is expressed in SI base units, and the product pV must come out in joules. One pascal is one newton per square metre, so Pa × m³ gives N m, which is a joule. Using dm³ with pascals gives an answer 1000 times too large.
- 5
Calculate the volume occupied by 0.0100 mol of an ideal gas at 101 000 Pa and 298 K. Give your answer in m³ and in cm³. (R = 8.31 J K⁻¹ mol⁻¹)
Model answer
V = nRT/p = (0.0100 × 8.31 × 298) ÷ 101 000 = 24.76 ÷ 101 000 = 2.45 × 10⁻⁴ m³. Since 1 m³ = 1 × 10⁶ cm³, this is 245 cm³. The data are given to three significant figures, so the answer is given to three.
- 6
Deduce the units of the equilibrium constant Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Model answer
Kc = [NH₃]² ÷ ([N₂][H₂]³). Substituting mol dm⁻³ for every concentration gives (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻², so the units are mol⁻² dm⁶. The units must be derived from the expression each time; they are not fixed for all equilibria.
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