In the Dumas method for the estimation of nit | vedclass

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(A) $1$. Pressure of dry $N_2$ gas $(P_{N_2})$ = $P_{total} - P_{aqueous} = 746 \ mm - 6 \ mm = 740 \ mm$. $2$. Convert pressure to $atm$: $P = 740 /

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$0.39$

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(A) $1$. Pressure of dry $N_2$ gas $(P_{N_2})$ = $P_{total} - P_{aqueous} = 746 \ mm - 6 \ mm = 740 \ mm$. $2$. Convert pressure to $atm$: $P = 740 / 760 \ atm$. $3$. Volume of $N_2$ at $STP$ $(V_0)$: Using the ideal gas law $\frac{P_1 V_1}{T_1} = \frac{P_0 V_0}{T_0}$,where $P_0 = 760 \ mm$,$T_0 = 273 \ K$,$V_1 = 1.31 \ mL$,$P_1 = 740 \ mm$,$T_1 = 293 \ K$. $V_0 = \frac{740 	imes 1.31 	imes 273}{760 	imes 293} \approx 1.216 \ mL$. $4$. Mass of $N_2$ = $\frac{28 	imes V_0 (in \ L)}{22400} = \frac{28 	imes 1.216}{22400} \approx 0.00152 \ g$. $5$. Percentage of nitrogen = $\frac{	ext{Mass of } N_2}{	ext{Mass of compound}} 	imes 100 = \frac{0.00152}{0.388} 	imes 100 \approx 0.39 \%$.

In the Dumas method for the estimation of nitrogen,$0.388 \ g$ of an organic compound yielded $1.31 \ mL$ of $N_2$ gas collected at $293 \ K$ temperature and $746 \ mm$ pressure. Calculate the percentage of nitrogen in the compound. (Aqueous tension at $293 \ K = 6 \ mm$). (in $\%$)

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