In Duma's method of estimation of nitrogen,0. | vedclass

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(B) First,calculate the pressure of dry nitrogen gas by subtracting the aqueous tension from the total pressure: $P_{N_2} = 715 \ mm - 15 \ mm = 700 \

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$16$

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(B) First,calculate the pressure of dry nitrogen gas by subtracting the aqueous tension from the total pressure: $P_{N_2} = 715 \ mm - 15 \ mm = 700 \ mm \ Hg$. Next,convert the volume of nitrogen to $STP$ conditions using the ideal gas equation: $V_{STP} = \frac{V 	imes P 	imes 273}{T 	imes 760} = \frac{55 	imes 700 	imes 273}{300 	imes 760} \approx 46.05 \ mL$. Finally,calculate the percentage of nitrogen using the formula: $	ext{Percentage of } N = \frac{28}{22400} 	imes \frac{V_{STP}}{	ext{mass of compound}} 	imes 100$. Substituting the values: $	ext{Percentage of } N = \frac{28}{22400} 	imes \frac{46.05}{0.35} 	imes 100 \approx 16.45 \%$. Thus,the percentage of nitrogen in the compound is $16.45 \%$.

In $Duma's$ method of estimation of nitrogen,$0.35 \ g$ of an organic compound gave $55 \ mL$ of nitrogen collected at $300 \ K$ temperature and $715 \ mm$ pressure. The percentage of nitrogen in the compound is (aqueous tension at $300 \ K = 15 \ mm$). (in $.45$)

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