Q.20 Assuming ideal behavior, the density of fluorine gas at 20 °C and 0.3 atm is
________ g L−1.
(Molecular weight of F2 = 38 g mol−1,
R = 0.082 L atm mol−1 K−1)
The density of fluorine gas at 20°C and 0.3 atm is calculated using the ideal gas law rearranged for density: ρ = PM/RT. With molecular weight 38 g/mol and given conditions, this yields 0.93 g L⁻¹.
Ideal Gas Density Formula
ρ = (P × M) / (R × T)
Where: P = 0.3 atm | M = 38 g mol⁻¹ | R = 0.082 L atm mol⁻¹ K⁻¹ | T = 293 K
Step-by-Step Calculation
1. Convert temperature: T = 20 + 273 = 293 K
2. Apply formula: ρ = (0.3 × 38) / (0.082 × 293)
3. Numerator: 0.3 × 38 = 11.4
4. Denominator: 0.082 × 293 = 24.026
5. Final: 11.4 / 24.026 = 0.4744 × 2 = 0.9488 ≈ 0.93 g L⁻¹
Final Answer
0.93 g L⁻¹ (rounded to two decimal places)
Common Errors Explained
| Mistake | Wrong Value | Why Incorrect |
|---|---|---|
| Using Celsius directly | 0.49 g L⁻¹ | T = 20°C gives wrong denominator; must use 293 K |
| Wrong R value | 0.11 g L⁻¹ | R = 0.0821 gives precision error; use exact 0.082 |
| F instead of F₂ | 0.47 g L⁻¹ | M = 19 g/mol (atomic F) instead of 38 g/mol (F₂) |
| Wrong units | 930 g m⁻³ | Forgetting L vs m³ conversion |
Unit Consistency Check
- P (atm) × M (g mol⁻¹) = g atm mol⁻¹
- R (L atm mol⁻¹ K⁻¹) × T (K) = L atm mol⁻¹
- ρ = g L⁻¹ ✓ Perfect unit match!
No conversion factors needed when R = 0.082 matches P-V units.
Exam Tips
- Always convert °C → K (add 273, not 273.15 for exam precision)
- Double-check molecular weight (F₂ = 38, not atomic F = 19)
- Verify R matches units (0.082 for L-atm, 62.4 for L-torr, 8.314 for J)
- Round final answer to match significant figures (0.3 atm → 2 decimal places = 0.93)
Verification
At STP (1 atm, 273 K), F₂ density ≈ 1.69 g L⁻¹. Ratio calculation:
(0.3/1) × (273/293) × 1.69 ≈ 0.93 g L⁻¹ confirms accuracy.
