Q88 / 200 · NEET 2011 (Screening) · Chemistry
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 $300 \mathrm{~K}=15 \mathrm{~mm}$ )
- Correct answerA16.45
- B17.45
- C14.45
- D15.45
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
$$
\therefore V_2=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 \mathrm{~mL}
$$
$$
\begin{aligned}
& \text { \% of nitrogen in given compound } \\
& =\frac{28}{22400} \times \frac{V_2}{W} \times 100=\frac{28}{22400} \times \frac{46.098}{0.35} \times 100 \\
& =16.45 \%
\end{aligned}
$$
Hence, the correct answer is option: 'A'.