In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given: Aqueous tension at 300 K = 15 mm Hg)

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Solution

Q.
In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is

(Given: Aqueous tension at 300 K = 15 mm Hg) [2025]

1 15.71%

2 20.95%

3 17.46%

4 7.85%

Ans.
(1)

Pressure of nitrogen $(P_{N_{2}})$ = Total pressure - aqueous tension

$P_{N_{2}} = (715 - 15) mmHg = 700 mmHg = \frac{700}{760} atm$

Volume of nitrogen $(V_{N_{2}})$ = 60 mL = 0.06 L

Moles of nitrogen $(n_{N_{2}}) = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{\frac{700}{760} \times 0.06}{0.0821 \times 300}$

Mass of nitrogen $(W_{N_{2}}) = n_{N_{2}} \times$ molar mass of nitrogen

$= \frac{\frac{700}{760} \times 0.06}{0.0821 \times 300} \times 28 g$

Mass percentage of nitrogen

$= \frac{W_{N_{2}}}{Mass of organic compound} \times 100$

$= \frac{\frac{700}{760} \times 0.06 \times 28}{0.0821 \times 300 \times 0.4} \times 100 = 15.71 %$

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