Summary AQA AS Level Physical Chemistry - Unit 3.1.2 - Amount of Substance - Full Notes
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Course
Unit 3.1.2 - Amount of substance (74041)
Institution
AQA
Detailed notes for AQA AS level chemistry unit 3.1.2 (amount of substance). Covers: molecular/empirical formulae, mole/mass/Mr/Ar equations, ideal gas equation, percentage error/uncertainty, percentage yield, atom economy. Includes example calculations. Also covers methodology for (1) preparing a s...
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3.1.2.1 Can you describe relative atomic mass and relative molecular mass in terms of C?
Can you define relative atomic mass (Ar)?
Can you define relative molecular mass (Mr)?
Relative Atomic Mass (Ar): The average mass of an element (taking into account all of its isotopes) relative to 1/12 the
mass of a 12C atom.
3.1.2.2 Can you define the Avogadro constant as the number of particles in a mole?
Can you describe the mole as applied to electrons, atoms, molecules, ions, formulae and equations?
Can you use the Avogadro constant to carry out calculations?
Can you use mass of substance, Mr, and amount in moles to carry out calculations?
• One mole contains 6.022 x 1023 (L) atoms/ions/etc
Moles = Mass (g) n = _m_ Mr = Mass (g) x L
Mr Mr
3.1.2.2 Can you measure the concentration of a substance in solution, measured in mol dm–3?
Moles (mol) = Concentration (mol/dm-3) x Volume (dm3) n = cv
Mass (g) = Concentration (g/dm-3) x Volume (dm3)
Concentration (g/dm-3) = Concentration (mol/dm-3) x Mr
For two solutions with the same number of moles: C1V1 = C2V2
, 3.1.2.3 Do you know the ideal gas equation pV = nRT with the variables in SI units?
Can you use the ideal gas equation pV = nRT in calculations?
Avogadro’s Law states that equal volumes of gas at the same temperature and pressure contain equal numbers of
molecules.
1 mole of any gas occupies 24dm3 at room temperature and pressure.
Calculate the relative molecular mass of a gas which has a density of 2.615
g dm-3 at 298K and 1 atm. PV = nRT
P = 1 atm = 101 kPa = 1.01x105 Pa n = _P_
V RT
R = 8.31 Jk-1mol-1
T = 298K _m_ = _P_
ρ = 2.615 g dm-3 = 2615 g m-3 Mr V RT
_ρ_ = _P_
Mr RT
Mr = ρRT
P
Mr = 2615 x 8.31 x 298
1.01x105
Mr = 64.1 (3sf)
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