JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ______ %

(Aqueous tension at 300 K is 15 mm).

Your answer

Solution

Correct $N_{2}$ pressure (subtract aqueous tension):

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

Volume: $V = 70$ mL $= 0.070$ L.

Moles of $N_{2}$ from ideal gas law:

$n_{N_{2}} = \frac{P V}{R T} = \frac{( 700/760 ) ( 0.070 )}{0.0821 \times 300}$ .

Mass of $N_{2}$ (multiplying by molar mass 28):

$W_{N_{2}} = n_{N_{2}} \times 28 = \frac{700 \times 0.070 \times 28}{760 \times 0.0821 \times 300} = \frac{1372}{18717.5} \approx 0.0733$ g.

$% N = \frac{0.0733}{0.50} \times 100 \approx 14.65% \approx 15%$ .

Answer: $15$ .

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