JEE Main 2025 Apr 3 Shift 2, Chemistry Q11: Mole Concept And Equivalent Concept

JEE Main2025Apr 3, Shift 2Chemistry

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)

A
15.71%
B
20.95%
C
17.46%
D
7.85%

Solution

The N $_{2}$ gas is collected over water, so the dry-gas pressure is the total pressure minus the aqueous tension:

$P_{N_{2}} = 715 - 15 = 700$ mm Hg $= \frac{700}{760}$ atm.

Apply the ideal gas law to get moles of N $_{2}$ (with V = 0.060 L, T = 300 K):

$n_{N_{2}} = \frac{P V}{R T} = \frac{( 700/760 ) ( 0.060 )}{0.0821 \times 300} \approx 0.00225$ mol.

Mass of N in the sample $= 2 \times 14 \times n_{N_{2}} = 28 \times 0.00225 \approx 0.063$ g.

$% N = \frac{0.063}{0.4} \times 100 \approx 15.71%$ .

$∴$ The answer is 15.71%.

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JEE Main 2025 Apr 3 Shift 2, Chemistry Q11: Mole Concept And Equivalent Concept

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