A sample of 0.125 ~g of an organic compound when analyzed 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 \underline\hspace2cm . (Nearest integer) Given : (a) The vapour pressure of water of 280 ~K is 14.2 ~mm \,Hg . (b) R=0.082 ~L atm K^-1 ~mol^-1

Practical Organic Chemistry PYQ — JEE Main 2022 · 28 Jul · Shift 2 · Q18 with Solution

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Correct answer: 22

Mass of organic compound, $m = 0.125\,\text{g}$
Volume of nitrogen gas, $V = 22.78\,\text{mL} = 0.02278\,\text{L}$
Temperature, $T = 280\,\text{K}$
Total pressure, $P_{\text{total}} = 759\,\text{mm Hg}$
Vapour pressure of water at $280\,\text{K}$ , $P_{\ce{H2O}} = 14.2\,\text{mm Hg}$
Gas constant, $R = 0.082\,\text{L atm K}^{-1}\text{mol}^{-1}$

Since nitrogen is collected over KOH, $\ce{CO2}$ is absorbed, but the gas is still moist. So pressure of dry nitrogen is:

P_{\ce{N2}} = P_{\text{total}} - P_{\ce{H2O}} = 759 - 14.2 = 744.8\,\text{mm Hg}

Convert into atm:

P_{\ce{N2}} = \frac{744.8}{760} = 0.98\,\text{atm}

PV = nRT

So,

n_{\ce{N2}} = \frac{PV}{RT} = \frac{0.98 \times 0.02278}{0.082 \times 280}

First, numerator:

0.98 \times 0.02278 = 0.0223244

Denominator:

0.082 \times 280 = 22.96

Thus,

n_{\ce{N2}} = \frac{0.0223244}{22.96} \approx 9.72 \times 10^{-4}\,\text{mol}

Each mole of $\ce{N2}$ has mass $28\,\text{g}$ .

\text{Mass of nitrogen} = n_{\ce{N2}} \times 28

= 9.72 \times 10^{-4} \times 28 \approx 0.0272\,\text{g}

\%\text{N} = \frac{\text{mass of nitrogen}}{\text{mass of compound}} \times 100

\%\text{N} = \frac{0.0272}{0.125} \times 100

\%\text{N} = 21.76\%

Nearest integer:

\boxed{22}

Stored correct answer = $22$

Our derived answer also = $22$ . So they agree.