In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _______ % (nearest integer).
(Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )
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Correct Answer: 20
Approach Solution - 1
To find the percentage composition of nitrogen in the compound using Dumas' method, follow these steps:
Step 1: Calculate the number of moles of nitrogen gas.
We use the ideal gas law: PV = nRT.
P (pressure) = 900 mm Hg - 15 mm Hg (aqueous tension) = 885 mm Hg.
Convert pressure to atm: P = 885 mm Hg × (1 atm / 760 mm Hg) ≈ 1.164 atm.
V (volume) = 150 mL = 0.150 L.
R (gas constant) = 0.0821 L·atm/(mol·K).
T (temperature) = 300 K.
Plug these values into the ideal gas equation: n = (1.164 atm × 0.150 L) / (0.0821 L·atm/(mol·K) × 300 K).
Calculate n, the number of moles of nitrogen: n ≈ 0.0071 mol.
Step 2: Calculate the mass of nitrogen.
Molar mass of nitrogen (N2) = 28.02 g/mol.
Mass of nitrogen = 0.0071 mol × 28.02 g/mol ≈ 0.198762 g.
Step 3: Calculate the percentage composition of nitrogen.
Percentage of nitrogen = (mass of nitrogen / mass of compound) × 100%.
Percentage of nitrogen = (0.198762 g / 1 g) × 100% ≈ 19.8762%.
Thus, the percentage composition of nitrogen is approximately 20%.
Conclusion: The calculated percentage composition of nitrogen aligns with the expected range of 20%.
Approach Solution -2
2. Calculate the moles of $\mathrm{N}_{2}$: \[ \text{Moles of } \mathrm{N}_2 = \frac{885 \mathrm{~mm~Hg} \times 0.15 \mathrm{~L}}{0.0821 \mathrm{~L~atm/mol} \times 300 \mathrm{~K}} = 0.0071 \mathrm{~mol} \]
3. Calculate the percentage of nitrogen: \[ \text{Percentage of nitrogen} = \frac{0.0071 \mathrm{~mol} \times 28 \mathrm{~g/mol}}{1 \mathrm{~g}} \times 100 = 19.85% \approx 20% \] Therefore, the correct answer is (20).
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