$0.1840\, gm$ of organic compound gave $30 \,mL$ of nitrogen which is collected at $287\, K$ And $758\,mm$ of $Hg$.

Given ; Aqueous tension at $287 \,K =14 \,mm$ of $Hg$. Hence actual pressure $=(758-14)$

$=744\, mm \text { of } Hg \text { . }$

Volume of nitrogen at $STP =\frac{273 \times 744 \times 30}{287 \times 760}$

$V =27.935 \,mL$

$\because 22400 mL$ of $N _{2}$ at STP weighs $=28\, gm .$

$\therefore 27.94 mL$ of $N _{2}$ at STP weighs $=$

$\left(\frac{28}{22400} \times 27.94\right)\, gm$

$=0.0349 \,gm$

Hence $=\left(\frac{0.0349}{0.1840} \times 100\right)$

$=18.97\, \%$

Rond off. Answer $=19 \,\%$