\(0.1840\, gm\) 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 \(=19 \,\%\)