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(B) Step $1$: Calculate the pressure of dry nitrogen gas. $P_{N_2} = P_{total} - P_{aqueous\ tension} = 715 \ mm - 15 \ mm = 700 \ mm$. Step $2$: Conv

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$17.46$

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(B) Step $1$: Calculate the pressure of dry nitrogen gas. $P_{N_2} = P_{total} - P_{aqueous\ tension} = 715 \ mm - 15 \ mm = 700 \ mm$. Step $2$: Convert the volume of nitrogen to $STP$ conditions using the ideal gas equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$. $P_1 = 700 \ mm, V_1 = 50 \ mL, T_1 = 300 \ K$. $P_2 = 760 \ mm, T_2 = 273 \ K$. $V_2 = \frac{700 	imes 50 	imes 273}{760 	imes 300} = 42.006 \ mL$. Step $3$: Calculate the mass of nitrogen. $22400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$. Mass of $N_2 = \frac{28 	imes 42.006}{22400} = 0.0525 \ g$. Step $4$: Calculate the percentage of nitrogen. $\% \ N = \frac{	ext{mass of } N_2}{	ext{mass of compound}} 	imes 100 = \frac{0.0525}{0.3} 	imes 100 = 17.5 \%$. The closest option is $17.46 \%$.

$0.3 \ g$ of an organic compound gave $50 \ mL$ of nitrogen collected at $300 \ K$ and $715 \ mm$ pressure in Dumas method. Calculate the percentage of nitrogen in the compound (aqueous tension of water at $300 \ K$ is $15 \ mm$).

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