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Practical Organic Chemistry question

2022 · 24 Jun · Shift 2 · Q19
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Practical Organic Chemistry question

2022 · 24 Jun · Shift 2 · Q19

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N2N_2N2​ evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is ‾\underline{\hspace{2cm}}​. [nearest integer] (Given : Molar mass of N2N_2N2​ is 28 g mol −-− 1. Molar volume of N2N_2N2​ at STP : 22.4 L)
Numerical answer
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Correct answer: 14

  1. Given data
  • Mass of organic compound =0.2 g= 0.2\,\text{g}=0.2g
  • Volume of N2N_2N2​ at STP =22.400 mL= 22.400\,\text{mL}=22.400mL
  • Molar volume at STP =22.4 L=22400 mL= 22.4\,\text{L} = 22400\,\text{mL}=22.4L=22400mL
  • Molar mass of N2=28 g mol−1N_2 = 28\,\text{g mol}^{-1}N2​=28g mol−1
  1. Find moles of N2N_2N2​ evolved

At STP,

22400 mL of N2=1 mol22400\,\text{mL of } N_2 = 1\,\text{mol}22400mL of N2​=1mol

So,

moles of N2=22.40022400=0.001 mol\text{moles of } N_2 = \frac{22.400}{22400} = 0.001\,\text{mol}moles of N2​=2240022.400​=0.001mol
  1. Find mass of nitrogen gas obtained
mass of N2=0.001×28=0.028 g\text{mass of } N_2 = 0.001 \times 28 = 0.028\,\text{g}mass of N2​=0.001×28=0.028g

This entire nitrogen came from the organic compound.

  1. Calculate percentage of nitrogen
%N=mass of nitrogenmass of compound×100\%N = \frac{\text{mass of nitrogen}}{\text{mass of compound}} \times 100%N=mass of compoundmass of nitrogen​×100 %N=0.0280.2×100=14\%N = \frac{0.028}{0.2} \times 100 = 14%N=0.20.028​×100=14
  1. Final answer

The percentage of nitrogen in the compound is:

14\boxed{14}14​
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