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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V =22.78 ml , T =280 K P text total =759 mm Hg P N 2=759-14.2=744.8 mm Hg n N 2=(744.8 × 22.78/760 × 1000 × 0.082 × 280)=0.00097 W text Nitrogen =0.02716 % N =(0.02716/0.125) × 1000=21.728

V=22.78ml,T=280K Ptotal ​=759mmHg PN2​​=759−14.2=744.8mmHg nN2​​=760×1000×0.082×280744.8×22.78​=0.00097 WNitrogen ​=0.02716 %N=0.1250.02716​×1000=21.728

$V = 22.78 m l, T = 280 K$
$P_{total} = 759 mm H g$
$P_{N_{2}} = 759 - 14.2 = 744.8 mm H g$
$n_{N_{2}} = \frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280} = 0.00097$
$W_{Nitrogen} = 0.02716$
$% N = \frac{0.02716}{0.125} \times 1000 = 21.728$