In Duma's method for the estimation of nitrogen $0.25 g$ of an organic compound gave fumes of nitrogen collected at $300 K$ temperature and $725 mm$ pressure, if the aqueous tension at $300 K$ is $25 mm$ the percentage of nitrogen in the compound is?

15.76

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16.80

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17.36

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18.20

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Solution

The correct option is B $16.80$
$P_{d r y g a s} = P_{m o i s t g a s} - P_{A q . t e n s i o n}$
$\Rightarrow 725 - 25 = 700 mm$
$W_{N_{2}} = \frac{P V M}{R T} = \frac{700 \times 40 \times 28}{760 \times 1000 \times 300 \times 0.0821}$
$\Rightarrow W_{N_{2}} = 0.042 g$
$\Rightarrow % N_{2} = \frac{0.042}{0.25} \times 100 = 16.80 %$

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In Duma's method for the estimation of nitrogen 0.25 g of an organic compound gave fumes of nitrogen collected at 300 K temperature and 725 mm pressure, if the aqueous tension at 300 K is 25 mm the percentage of nitrogen in the compound is?

The correct option is B 16.80Pdry gas=Pmoist gas−PAq. tension ⇒725−25=700 mm WN2=PVMRT=700×40×28760×1000×300×0.0821 ⇒WN2=0.042 g ⇒%N2=0.0420.25×100=16.80 %

In Duma's method for the estimation of nitrogen $0.25 g$ of an organic compound gave fumes of nitrogen collected at $300 K$ temperature and $725 mm$ pressure, if the aqueous tension at $300 K$ is $25 mm$ the percentage of nitrogen in the compound is?