Chemistry - PRACTICAL ORGANIC CHEMISTRY Question with Solution | TestHub

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$ Where $p_2=$ pressure of ${\mathrm{N}}_2$ at $STP=760\mathrm{mm}$ $$\begin{array}{rl}& T_2=\text{ Temperature of }{\mathrm{N}}_2\text{ at }\mathrm{STP}=273\mathrm{K}\\ & V_2=\text{ ? }\end{array}$$ Volume of ${\mathrm{N}}_2$ at STP (By gas equation) $$\left(\frac{\rho -{\rho }_1}{t+273}\right)V_1\times \frac{273}{760}=V_2$$ where $p_1=\rho -{\rho }_1$ $\rho =7/5\mathrm{mm}$ (pressure at which ${\mathrm{N}}_2$ collected). ${\rho }_1=$ aqueous tension of water $=15\mathrm{mm}$ $T_1=t+273=300\mathrm{K}$ $V_1=55\mathrm{mL}=$ volume of moist nitrogen in nitrometer $$\therefore V_2=\frac{(715-15)\times 55}{300}\times \frac{273}{760}=46.098\mathrm{mL}$$ $$\begin{array}{rl}& \text{ \% of nitrogen in given compound }\\ & =\frac{28}{22400}\times \frac{V_2}{W}\times 100=\frac{28}{22400}\times \frac{46.098}{0.35}\times 100\\ & =16.45\%\end{array}$$