understand that amount of substance is an SI base quantity with the base unit mol

• The ideal‑gas law assumes point‑like particles with no intermolecular forces.
• It works well at low pressure and high temperature.
• At high pressure or low temperature real gases deviate; corrections are given by the Van der Waals equation.

Goal: Determine the molar mass of a volatile liquid and, from it, an experimental value of \(N_{\!A}\).

Checklist for Planning & Evaluation

1. Apparatus
   • Analytical balance (±0.001 g)
   • Heated bulb or evaporating dish
   • Thermometer, barometer
   • Glassware for collecting vapour (e.g., eudiometer)
2. Procedure (summary)
   1. Weigh the empty bulb (mass \(m_0\)).
   2. Add a known mass of liquid (mass \(m_{\text{liq}}\)).
   3. Heat gently until the liquid vaporises completely and the vapour fills the bulb.
   4. Cool the bulb, then weigh the bulb with vapour (mass \(m_1\)).
   5. Record temperature \(T\) and pressure \(p\) of the vapour.
3. Calculations
   • Mass of vapour: \(m_{\text{vap}} = m_1 - m_0\).
   • Density of vapour: \(\rho_{\text{vap}} = \dfrac{m_{\text{vap}}}{V_{\text{bulb}}}\) (bulb volume known).
   • Use the ideal‑gas law to find the molar mass: \[ M = \frac{\rho_{\text{vap}}\,RT}{p} \]
   • Experimental Avogadro number: \[ N_{\!A}^{\text{exp}} = \frac{M}{m_{\text{u}}} \] where \(m_{\text{u}} = 1.66053906660\times10^{-24}\ \text{g}\) (atomic mass unit).
4. Error analysis
   • Random errors: balance reading, temperature fluctuations.
   • Systematic errors: incomplete vapourisation, heat loss, leakage.
   • Propagate uncertainties through the molar‑mass formula (use partial‑derivative method or percentage‑error approximation).
5. Evaluation (Paper 5 style)
   • Discuss how well the assumptions of the ideal‑gas law are satisfied.
   • Identify the dominant source of error and suggest improvements (e.g., using a thermostatted bath, more accurate volume measurement).

• Scientific notation & significant figures – e.g., \(6.02214076\times10^{23}\) (exact) vs. \(6.02\times10^{23}\) (3 sf).
• Algebraic rearrangement – from \(pV=nRT\) obtain \(p=\dfrac{nRT}{V}\), \(V=\dfrac{nRT}{p}\), \(T=\dfrac{pV}{nR}\).
• Proportional‑reasoning exercise
  If 0.5 mol of an ideal gas occupies 12.0 L at 273 K and 1.00 atm, what pressure will 1.0 mol occupy at the same temperature if the volume is reduced to 6.0 L?
  Solution:
  \(p_1V_1=n_1RT\) → \(p_1 =\dfrac{n_1RT}{V_1}\).
  Because \(n\) and \(T\) are unchanged, \(p\propto\frac{1}{V}\); halving the volume doubles the pressure: \(p_1 = 2.00\ \text{atm}\).
• Scientific‑notation manipulation activity
  Write \(6.022\times10^{23}\) to 8 significant figures and then back to 3 significant figures.
  Answer: \(6.0221408\times10^{23}\) → \(6.02\times10^{23}\).
• Error propagation (basic) – for \(M=\dfrac{\rho RT}{p}\), the relative uncertainty is \[ \frac{\Delta M}{M}= \sqrt{\left(\frac{\Delta\rho}{\rho}\right)^2+\left(\frac{\Delta T}{T}\right)^2+\left(\frac{\Delta p}{p}\right)^2 }. \]

• Models & testing: The mole is a model that quantifies a collection of particles; experiments such as the vapour‑density method test the model’s validity.
• Matter & energy: Molar mass links the mass of a macroscopic sample to the energy per particle in chemical reactions.
• Mathematics: Linear relationships in the ideal‑gas law demonstrate how mathematics translates a physical model into quantitative predictions.