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

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(B) Given: Mass of organic compound $(m)$ = $0.25 \ g$,Volume of $N_2$ $(V_1)$ = $40 \ mL$,Temperature $(T_1)$ = $300 \ K$,Pressure $(P_1)$ = $725 \ m

Vedclass · Accepted Answer

$16.76$

Vedclass · Answer

(B) Given: Mass of organic compound $(m)$ = $0.25 \ g$,Volume of $N_2$ $(V_1)$ = $40 \ mL$,Temperature $(T_1)$ = $300 \ K$,Pressure $(P_1)$ = $725 \ mm - 25 \ mm = 700 \ mm$ (corrected pressure).Using the ideal gas equation at $STP$ $(P_0 = 760 \ mm, T_0 = 273 \ K)$:$V_0 = \frac{P_1 V_1}{T_1} 	imes \frac{T_0}{P_0} = \frac{700 \ mm 	imes 40 \ mL}{300 \ K} 	imes \frac{273 \ K}{760 \ mm} = 33.53 \ mL$.At $STP$,$22400 \ mL$ of $N_2$ weighs $28 \ g$.Mass of $N_2 = \frac{28 \ g 	imes 33.53 \ mL}{22400 \ mL} = 0.0419 \ g$.Percentage of nitrogen = $\frac{	ext{Mass of } N_2}{	ext{Mass of compound}} 	imes 100 = \frac{0.0419 \ g}{0.25 \ g} 	imes 100 = 16.76 \%$.Thus,the correct option is $B$.

In $Duma's$ method for estimation of nitrogen,$0.25 \ g$ of an organic compound gave $40 \ mL$ 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:

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