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JIPMER 2014
Exam
JIPMER
Year
2014
Subject
Chemistry
Chapter
General Organic Chemistry

JIPMER 2014 · Chemistry · General Organic Chemistry

In Duma's method of estimation of nitrogen, 0.35 g of an organic...

In Duma's method of estimation of nitrogen, 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K=15 mm300 \mathrm{~K}=15 \mathrm{~mm} )

Options

  1. A16.4516.45
  2. B17.4517.45
  3. C14.4514.45
  4. D15.4515.45

Correct answer

A. 16.4516.45

Step-by-step solution

p1V1T1=p2V2T2 or  (ρ−ρ1)T1V1=p2V2T2\begin{aligned} & \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \\ & \text{\text{ or } } \frac{\left(\rho-\rho_1\right)}{T_1} V_1=\frac{p_2 V_2}{T_2}\end{aligned} where, p1=ρ−ρ1ρ=715 mmp_1=\rho-\rho_1\rho=715 \mathrm{~mm} (pressure at which N2\mathrm{N}_2 collected). ρ1= aqueous tension of  water =15 mmt1=t+273=300 KV1=55 mL= Volume of moist nitrogen in  nitrometer ρ2= pressure of N2 at STP=760 mmt2= temperature of N2 at STP=273 KV2= ? \begin{aligned} & \rho_1=\text{ aqueous tension\text{ of } water }=15 \mathrm{~mm} \\ & t_1=t+273=300 \mathrm{~K} \\ & V_1=55 \mathrm{~mL}=\text{ Volume of moist nitrogen in } \\ & \text{ nitrometer } \\ & \rho_2=\text{ pressure of } \mathrm{N}_2 \text{ at } \mathrm{STP}=760 \mathrm{~mm} \\ & t_2=\text{ temperature of } \mathrm{N}_2 \text{ at } \mathrm{STP}=273 \mathrm{~K} \\ & V_2=\text{ ? }\end{aligned}\therefore V_2=\frac{(715-15)}{300} \times 55 \times \frac{273}{760}=46.098 \mathrm{~mL} \%ofnitrogeninthegivencompoundof nitrogen in the given compound=2822400×V2W×100=2822400×46.0980.35×100=16.45%\begin{aligned} & =\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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