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Practical Organic Chemistry question

2021 · 16 Mar · Shift 2 · Q20
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Practical Organic Chemistry question

2021 · 16 Mar · Shift 2 · Q20

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]
Numerical answer
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Correct answer: 19

  1. Correct the pressure of dry nitrogen

Nitrogen is collected over water, so pressure of dry N2N_2N2​ is:

PN2=758−14=744 mm HgP_{N_2} = 758 - 14 = 744\ \text{mm Hg}PN2​​=758−14=744 mm Hg

Convert to atm:

PN2=744760 atmP_{N_2} = \frac{744}{760} \text{ atm}PN2​​=760744​ atm

  1. Use the ideal gas equation to find moles of nitrogen

Given:

  • Volume of N2=30 mL=0.030 LN_2 = 30\ \text{mL} = 0.030\ \text{L}N2​=30 mL=0.030 L
  • Temperature T=287 KT = 287\ \text{K}T=287 K
  • Gas constant R=0.0821 L atm mol−1K−1R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}R=0.0821 L atm mol−1K−1

Using

n=PVRTn = \frac{PV}{RT}n=RTPV​

n(N2)=(744760)(0.030)(0.0821)(287)n(N_2) = \frac{\left(\frac{744}{760}\right)(0.030)}{(0.0821)(287)}n(N2​)=(0.0821)(287)(760744​)(0.030)​

n(N2)≈0.9789×0.03023.5627n(N_2) \approx \frac{0.9789 \times 0.030}{23.5627}n(N2​)≈23.56270.9789×0.030​

n(N2)≈0.001246 moln(N_2) \approx 0.001246\ \text{mol}n(N2​)≈0.001246 mol

  1. Find mass of nitrogen present

Each mole of N2N_2N2​ has mass 28 g28\ \text{g}28 g.

Mass of nitrogen=0.001246×28\text{Mass of nitrogen} = 0.001246 \times 28Mass of nitrogen=0.001246×28

Mass of nitrogen≈0.0349 g\text{Mass of nitrogen} \approx 0.0349\ \text{g}Mass of nitrogen≈0.0349 g

  1. Calculate percentage of nitrogen

Mass of compound taken =0.1840 g= 0.1840\ \text{g}=0.1840 g

%N=0.03490.1840×100\%N = \frac{0.0349}{0.1840} \times 100%N=0.18400.0349​×100

%N≈18.97%\%N \approx 18.97\%%N≈18.97%

  1. Round to nearest integer

19\boxed{19}19​

  1. Comparison with stored answer

Derived answer = 191919

Stored correct answer = 191919

So, the answer matches the stored correct answer.

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