JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Practical and Analytical Chemistry

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In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is _____. (Round off to the nearest integer).
[Given: Aqueous tension at

287 K = 14 mm

of Hg]

Answer: 19

Solution

The nitrogen is collected over water, so the aqueous tension must be subtracted and the partial pressure of nitrogen is

758 - 14 = 744

mm Hg. With

V = 0.030

L and

T = 287

n = \frac{P V}{R T} = \frac{( 744/760 ) \times 0.030}{0.0821 \times 287} = 1.25 \times 1 0^{- 3} mol

which weighs

1.25 \times 1 0^{- 3} \times 28 = 0.0349

g. Against a sample of

0.1840

g this is

\frac{0.0349}{0.1840} \times 100 = 18.97

, so the compound contains 19 per cent nitrogen to the nearest integer.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Practical and Analytical Chemistry

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