In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N_2 ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ' N' in the organic compound is % (Near…

Question

In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N_2 ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ' N' in the organic compound is % (Nearest integer) (Aqueous tension at 300 K=15 mmHg )

Accepted Answer

Correct answer: 18. (18) Given: Volume of N_2 ,V=50 mL Pressure, P=715 mmHg Temperature, T=300 K Aqueous Tension at 300 K=15 mmHg Calculation: Correct the Pressure for N_2 : P_N_2=715 mmHg-15 mmHg=700 mmHg Convert the pressure from mm Hg to atm: P_N_2= 700760 atm Calculate the Moles of N_2 using the Ideal Gas Law: n_N_2= P_N_2 VR T n_N_2= 700760 5010000.0821 300 Calculate the Moles of N: n_N=2 n_N_2 Calculate the Mass of N: Mass of N=2 n_N 14 Determine the Percentage of Nitrogen in the Organic Compound: \%N= Mass ofN Mass of organic compound 100 \%N= 700760 501000 2 140.0821 300 1000292 100 \%N=18 \% The percentage of nitrogen in the organic compound is 18 \%.

$In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N_{2}) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ' N' in the organic compound is % (Nearest integer) (Aqueous tension at 300 K = 15 mmHg)$

Solution

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