In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )

<p><b>Step 1: Calculate the pressure of dry nitrogen gas</b></p> <p>The nitrogen gas was collected over water, so we subtract the aqueous tension (pressure of water vapor) from the total pressure to find the pressure of dry nitrogen gas.</p> <p>$$\text{Pressure of dry } \mathrm{N}_2 = 715\,\mathrm{mm\,Hg} - 15\,\mathrm{mm\,Hg} = 700\,\mathrm{mm\,Hg}$$</p> <p>Now, convert this pressure from mm Hg to atm:</p> <p>$700\,\mathrm{mm\,Hg} \div 760 = \frac{700}{760}\,\mathrm{atm}$</p> <p><b>Step 2: Use the ideal gas equation to find moles of nitrogen</b></p> <p>The volume of nitrogen ($V$) is given as 60 mL. Convert this to liters: 60 mL = 0.060 L (or use $60 \times 10^{-3}$ L). The temperature ($T$) is 300 K, and the gas constant ($R$) is 0.0821 L·atm·K⁻¹·mol⁻¹.</p> <p>Using the formula $\mathrm{n} = \frac{\mathrm{PV}}{\mathrm{RT}}$:</p> <p> $\mathrm{n} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300}$ </p> <p> When you calculate this, you get: $\mathrm{n} = 0.0022\;\text{mole}$ </p> <p><b>Step 3: Calculate the mass of nitrogen gas produced</b></p> <p>Moles of nitrogen are 0.0022. The molar mass of $\mathrm{N}_2$ is 28 g/mol.</p> <p> $\text{Mass of } \mathrm{N}_2 = 0.0022 \times 28 = 0.063\,\mathrm{g}$ </p> <p><b>Step 4: Find the percentage of nitrogen in the compound</b></p> <p>The mass of organic compound taken is 0.4 g. To find the percentage of nitrogen:</p> <p> $$ \text{Percentage of nitrogen} = \frac{\text{Mass of } \mathrm{N}_2}{\text{Mass of compound}} \times 100 $$ </p> <p> $= \frac{0.063}{0.4} \times 100 = 15.71\%$ </p>