In Duma's method of estimation of nitrogen… - NEET UG 2011 Chemistry PYQ

NEET UG 2011 — Chemistry Organic Chemistry

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In Duma's method of estimation of nitrogen $0.35 \mathrm{~g}$ of an organic compound gave $55 \mathrm{~mL}$ of nitrogen collected at $300 \mathrm{~K}$ temperature and $715 \mathrm{~mm}$ pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm}$ )

Official previous-year question

Held on 30 Apr 2011 · Verified 9 Jul 2026.

Options

A
16.45
B
17.45
C
14.45
D
15.45

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Solution

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$ Where $p_2=$ pressure of $\mathrm{N}_2$ at $S T P=760 \mathrm{~mm}$ $\begin{aligned} & T_2=\text { Temperature of } \mathrm{N}_2 \text { at } \mathrm{STP}=273 \mathrm{~K} \\ & V_2=\text { ? } \end{aligned}$ 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 $\therefore V_2=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 \mathrm{~mL}$ $\begin{aligned} & \text { \% of nitrogen in given compound } \\ & =\frac{28}{22400} \times \frac{V_2}{W} \times 100=\frac{28}{22400} \times \frac{46.098}{0.35} \times 100 \\ & =16.45 \% \end{aligned}$

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