Chemistry Q64 – JEE-MAIN 2025

Home Chemistry Organic Chemistry – Some Basic Principles and TechniquesQuestion #466

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 K = 15 mm Hg

)

Clue 1 of 3: Dumas' Method Principle

In Dumas' method, nitrogen gas is collected over an aqueous solution, so the observed pressure includes the vapor pressure of water (aqueous tension).

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Step 1: Calculate the pressure of dry nitrogen✦ Active

The observed pressure includes the aqueous tension. To find the pressure exerted by dry nitrogen, subtract the aqueous tension from the total observed pressure.

P_{N_{2}} = P_{o b s} - P_{a q}

Given $P_{o b s} = 715 mm Hg$ and $P_{a q} = 15 mm Hg$ :

P_{N_{2}} = 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 convert the volume of nitrogen from the given conditions ( $P_{1}, V_{1}, T_{1}$ ) to Standard Temperature and Pressure ( $P_{0} = 760 mm Hg$ , $T_{0} = 273 K$ ). The given conditions are $V_{1} = 60 mL$ , $P_{1} = 700 mm Hg$ , $T_{1} = 300 K$ .

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{0} V_{0}}{T_{0}} ⟹ V_{0} = V_{1} \times \frac{P_{1}}{P_{0}} \times \frac{T_{0}}{T_{1}}

Substitute the values:

V_{0} = 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 of nitrogen in the compound○ Expand

Use the formula for percentage of nitrogen in Dumas' method, which relates the volume of nitrogen at STP ( $V_{S T P}$ ) to the mass of the organic compound ( $W$ ). The molar volume of nitrogen at STP is $22400 mL/mol$ and its molar mass is $28 g/mol$ .

% N = \frac{28}{22400} \times \frac{V_{S T P}}{W} \times 100

Given $W = 0.4 g$ and $V_{S T P} \approx 50.289 mL$ :

% N = \frac{28}{22400} \times \frac{50.289 mL}{0.4 g} \times 100

% N = \frac{1}{800} \times \frac{50.289}{0.4} \times 100 \approx 15.715 %

Rounding to two decimal places, the percentage composition of nitrogen is $15.71 %$ . This matches option 2.

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Chemistry Question 64 – JEE-MAIN 2025

Clue 1 of 3: Dumas' Method Principle

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