JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Practical and Analytical Chemistry

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A sample of 0.125 g of an organic compound when analysed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg . The percentage of nitrogen in the given organic compound is ____. (Nearest integer).
(a) The vapour pressure of water at 280 K is 14.2 mm Hg
(b)

R = 0.082 L

atm

K^{- 1} mol^{- 1}

Answer: 22

Solution

Since nitrogen is collected over KOH solution (which absorbs CO₂ but not N₂), we need to correct for water vapor pressure.

Partial pressure of N₂:

P_{N_{2}} = 759 - 14.2 = 744.8

mm Hg =

\frac{744.8}{760}

atm

Using ideal gas law to find moles of N₂:

n_{N_{2}} = \frac{P V}{R T} = \frac{( 744.8/760 ) \times 0.02278}{0.082 \times 280} \approx 0.00097 mol

Mass of nitrogen:

m_{N} = 0.00097 \times 28 = 0.02716

g

Percentage of nitrogen:

% N = \frac{0.02716}{0.125} \times 100 = 21.73 \approx 22%

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Practical and Analytical Chemistry

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