Mole Concept – NEET PG Chemistry Notes
(Clear, concise, and exam‑ready)
1. Simple Explanation
| Term | What it Means |
|---|---|
| Mole | A unit of amount of substance. One mole contains the same number of entities as 12 g of carbon‑12 contains atoms. |
| Avogadro’s Number | (N_A = 6.022 \times 10^{23}) entities per mole. |
| Molar Mass | Mass of one mole of a substance (grams per mole). |
So, mass (g) ÷ molar mass (g mol⁻¹) = number of moles (mol).
And moles × Avogadro’s number = number of particles.
2. Key Points (NEET‑PG Focus)
| Concept | Formula | Notes |
|---|---|---|
| Avogadro's Constant | (N_A = 6.022 \times 10^{23}) | Constant, same for atoms, ions, molecules, formula units. |
| Molar Mass | (M = \frac{\text{mass of 1 mole}}{1}) | Use periodic table (g mol⁻¹). |
| Number of Moles | (n = \frac{m}{M}) | (m) = mass (g). |
| Number of Particles | (N = n \times N_A) | (N) may be atoms, ions, molecules, formula units. |
| Molar Volume of an Ideal Gas | (V_m = 22.414\ \text{L mol}^{-1}) at 273.15 K, 1 atm (STP). | Use (PV = nRT). |
| Stoichiometry | Use mole ratios from balanced equations. | Convert grams → moles → moles of product → grams of product. |
| Mole Ratio Example | (\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3) | Ratio (1 : 3 : 2). |
| Limiting Reactant | Compare actual mole ratio with stoichiometric ratio. | Determines maximum product. |
| Percent Yield | (\frac{\text{actual yield}}{\text{theoretical yield}} \times 100%) | Requires correct mole calculations. |
3. Worked Example (Typical NEET‑PG)
Problem
A sample of 5.00 g of sodium chloride (NaCl) is dissolved in water. If the solution has a density of 1.05 g mL⁻¹ and the molarity of NaCl is 0.800 M, calculate:
- The number of moles of NaCl present.
- The number of Na⁺ and Cl⁻ ions in the solution.
- The mass of Na⁺ and Cl⁻ ions separately.
Step‑by‑Step
| Step | Calculation | Result |
|---|---|---|
| 1. Molar mass of NaCl | (M_{\text{Na}} = 22.99), (M_{\text{Cl}} = 35.45) → (M_{\text{NaCl}} = 58.44\ \text{g mol}^{-1}) | |
| 2. Moles of NaCl | (n = \frac{m}{M} = \frac{5.00}{58.44} = 0.0856\ \text{mol}) | 0.0856 mol |
| 3. Number of Na⁺ and Cl⁻ ions | Each NaCl → 1 Na⁺ + 1 Cl⁻. | |
| (N = n \times N_A = 0.0856 \times 6.022\times10^{23} = 5.16\times10^{22}) ions | (5.16\times10^{22}\ \text{Na}^+) and (5.16\times10^{22}\ \text{Cl}^-) | |
| 4. Mass of Na⁺ | (m_{\text{Na}^+} = n \times M_{\text{Na}} = 0.0856 \times 22.99 = 1.97\ \text{g}) | 1.97 g |
| 5. Mass of Cl⁻ | (m_{\text{Cl}^-} = n \times M_{\text{Cl}} = 0.0856 \times 35.45 = 3.03\ \text{g}) | 3.03 g |
Check: (1.97 + 3.03 = 5.00) g (as given).
Quick Revision Summary
| Question | Answer |
|---|---|
| What is a mole? | Unit equal to Avogadro’s number of entities. |
| Avogadro’s Number? | (6.022\times10^{23}) |
| Molar Mass? | Grams per mole, derived from periodic table. |
| How to find moles? | (n = \frac{m}{M}) |
| How to find number of particles? | (N = n \times N_A) |
| Molar volume at STP? | 22.414 L mol⁻¹ |
| Stoichiometric mole ratio | Use balanced equation. |
| Limiting reactant determination | Compare actual to theoretical mole ratios. |
| Percent yield formula | (\frac{\text{actual}}{\text{theoretical}}\times100%) |
Remember: Always convert units first (g → mol, mol → ions, etc.), then apply the relevant formula.
Happy studying!