NEET UG 2011 — Chemistry Organic Chemistry

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$T1​p1​V1​​=T2​p2​V2​​ Where $p_2=$p2​= pressure of $\mathrm{N}_2$N2​ at $S T P=760 \mathrm{~mm}$STP=760 mm $$\begin{aligned} & T_2=\text { Temperature of } \mathrm{N}_2 \text { at } \mathrm{STP}=273 \mathrm{~K} \\ & V_2=\text { ? } \end{aligned}$$​T2​= Temperature of N2​ at STP=273 KV2​= ? ​ Volume of $\mathrm{N}_2$N2​ at STP (By gas equation) $$\left(\frac{\rho-\rho_1}{t+273}\right) V_1 \times \frac{273}{760}=V_2$$(t+273ρ−ρ1​​)V1​×760273​=V2​ where $p_1=\rho-\rho_1$p1​=ρ−ρ1​ $\rho=7 / 5 \mathrm{~mm}$ρ=7/5 mm (pressure at which $\mathrm{N}_2$N2​ collected). $\rho_1=$ρ1​= aqueous tension of water $=15 \mathrm{~mm}$=15 mm $T_1=t+273=300 \mathrm{~K}$T1​=t+273=300 K $V_1=55 \mathrm{~mL}=$V1​=55 mL= volume of moist nitrogen in nitrometer $$\therefore V_2=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 \mathrm{~mL}$$∴V2​=300(715−15)×55​×760273​=46.098 mL $$\begin{aligned} & \text { \% of nitrogen in given compound } \\ & =\frac{28}{22400} \times \frac{V_2}{W} \times 100=\frac{28}{22400} \times \frac{46.098}{0.35} \times 100 \\ & =16.45 \% \end{aligned}$$​ % of nitrogen in given compound =2240028​×WV2​​×100=2240028​×0.3546.098​×100=16.45%​