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NEET 2011
Exam
NEET
Year
2011
Subject
Chemistry
Chapter
General Organic Chemistry

NEET 2011 · 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 g0.35 \mathrm{~g} of an organic compound gave 55 mL55 \mathrm{~mL} of nitrogen collected at 300 K300 \mathrm{~K} temperature and 715 mm715 \mathrm{~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. A16.45
  2. B17.45
  3. C14.45
  4. D15.45

Correct answer

A. 16.45

Step-by-step solution

p1V1T1=p2V2T2\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} Where p2=p_2= pressure of N2\mathrm{N}_2 at STP=760mmS T P=760 \mathrm{ mm}T2= Temperature of  N2 at STP=273KV2= ? \begin{aligned}& T_2=\text{ Temperature\text{ of } } \mathrm{N}_{2} \text{ at } \mathrm{STP}=273 \mathrm{ K}\\ & V_2=\text{ ? } \end{aligned}VolumeofN2atSTP(Bygasequation)(ρ−ρ1t+273)V1×273760=V2whereVolume 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 mm(pressureatwhichN2collected).\mathrm{ mm} (pressure at which \mathrm{N}_2 collected).\rho_1=aqueoustensionofwateraqueous tension of water=15 mm\mathrm{ mm} T_1=t+273=300 K\mathrm{ K} V_1=55 mL=volume\mathrm{ mL}= volume of moistnitrogeninnitrometer∴V2=(715−15)×55300×273760=46.098mLmoist nitrogen in nitrometer \therefore V_2=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 \mathrm{ mL} % of nitrogen in given compound =2822400×V2W\ttimes100=2822400×46.0980.35×100=16.45%\begin{aligned}& \text{ \% of nitrogen in given compound }\\ & =\frac{28}{22400} \times \frac{V_2}{W} \t times100=\frac{28}{22400} \times \frac{46.098}{0.35} \times100 \\ & =16.45 \% \end{aligned}$$
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