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Practical Organic Chemistry question

2021 · 25 Jul · Shift 2 · Q22
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Practical Organic Chemistry question

2021 · 25 Jul · Shift 2 · Q22

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
0.8 g of an organic compound was analyzed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then ‾\underline{\hspace{2cm}}​ mL of 1 M H2SO4H_2SO_4H2​SO4​ would have been neutralized by the ammonia evolved during the analysis.
Numerical answer
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Correct answer: 12

  1. Mass of nitrogen in the compound

Given:

  • Mass of organic compound =0.8 g= 0.8\,\text{g}=0.8g
  • Percentage of nitrogen =42%= 42\%=42%

So, mass of nitrogen is

Mass of N=42100×0.8=0.336 g\text{Mass of N} = \frac{42}{100} \times 0.8 = 0.336\,\text{g}Mass of N=10042​×0.8=0.336g
  1. Moles of nitrogen atoms

Atomic mass of nitrogen =14 g mol−1= 14\,\text{g mol}^{-1}=14g mol−1

Thus,

Moles of N=0.33614=0.024 mol\text{Moles of N} = \frac{0.336}{14} = 0.024\,\text{mol}Moles of N=140.336​=0.024mol

In Kjeldahl's method, all nitrogen is converted into ammonia, so:

Moles of NH3=0.024 mol\text{Moles of } NH_3 = 0.024\,\text{mol}Moles of NH3​=0.024mol
  1. Reaction with sulfuric acid

Ammonia neutralizes sulfuric acid as:

2NH3+H2SO4→(NH4)2SO42NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_42NH3​+H2​SO4​→(NH4​)2​SO4​

From the equation:

  • 222 mol NH3NH_3NH3​ neutralize 111 mol H2SO4H_2SO_4H2​SO4​

Hence, moles of H2SO4H_2SO_4H2​SO4​ neutralized are

Moles of H2SO4=0.0242=0.012 mol\text{Moles of } H_2SO_4 = \frac{0.024}{2} = 0.012\,\text{mol}Moles of H2​SO4​=20.024​=0.012mol
  1. Volume of 1 M H2SO4H_2SO_4H2​SO4​

Using

M=nV(in L)M = \frac{n}{V(\text{in L})}M=V(in L)n​

so

V=nM=0.0121=0.012 LV = \frac{n}{M} = \frac{0.012}{1} = 0.012\,\text{L}V=Mn​=10.012​=0.012L

Convert to mL:

0.012 L=12 mL0.012\,\text{L} = 12\,\text{mL}0.012L=12mL
  1. Final answer
12\boxed{12}12​

The derived answer matches the stored correct answer.

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