Chemistry Q64 – JEE-MAIN 2025

Home Chemistry Organic Chemistry – Some Basic Principles and TechniquesQuestion #465

In Dumas' method for estimation of nitrogen, 0.5 gram 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 (Aqueous tension at 300 K = 15 mm Hg) is _______%.

Clue 1 of 3: Gas Pressure Correction

In Dumas' method, nitrogen is collected over water, so the observed pressure is the sum of the partial pressure of dry nitrogen and the aqueous tension.

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Step 1: Calculate the pressure of dry nitrogen✦ Active

The pressure of dry nitrogen ( $P_{N_{2}}$ ) is obtained by subtracting the aqueous tension from the total observed pressure.

P_{N_{2}} = P_{t o t a l} - P_{a q} = 715 mm Hg - 15 mm Hg = 700 mm Hg

Step 2: Calculate the volume of nitrogen at STP○ Expand

Use the combined gas law to find the volume of nitrogen ( $V_{S T P}$ ) at STP ( $P_{S T P} = 760 mm Hg$ , $T_{S T P} = 273 K$ ).

V_{S T P} = V_{e x p} \times \frac{P_{N_{2}}}{P_{S T P}} \times \frac{T_{S T P}}{T_{e x p}}

V_{S T P} = 60 mL \times \frac{700 mm Hg}{760 mm Hg} \times \frac{273 K}{300 K} \approx 50.289 mL

Step 3: Calculate the percentage composition of nitrogen○ Expand

The mass of nitrogen is calculated from its volume at STP using the molar volume ( $22400 mL$ at STP for $28 g$ of $N_{2}$ ). Then, the percentage is found relative to the mass of the organic compound.

% N = \frac{V_{S T P} (mL)}{22400 mL} \times 28 g \times \frac{1}{Mass of organic compound (g)} \times 100

Mass of N_{2} = \frac{50.289}{22400} \times 28 \approx 0.06286 g

Then, % N = \frac{0.06286 g}{0.5 g} \times 100 \approx 12.57 %

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Chemistry Question 64 – JEE-MAIN 2025

Clue 1 of 3: Gas Pressure Correction

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