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Question:
$\text{In Kjeldahl's method, the nitrogen present is estimated as-}$
Options:
  • 1. $\text{N}_2$
  • 2. $\text{NH}_3$ (Correct)
  • 3. $\text{NO}_2$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: Compound containing nitrogen is heated with concentrated sulphuric acid.}$ $\text{Complete Step by step solution:}$ $\text{Before we move forward with the solution of this given question, let us understand some basic concepts. The Kjeldahl method or Kjeldahi digestion in analytical chemistry is a method for the quantitative determination of nitrogen contained in organic substances plus the nitrogen contained in the inorganic compounds ammonia and ammonium.}$ $\text{Overall, the Kjeldahl method is divided into 3 main steps. The method has to be carried out in proper sequence. The steps include digestion, distillation, and titration.}$ $\text{1. Digestion: In this method, a certain substance or sample is heated in the presence of sulphuric acid. The acid breaks down the organic substance via oxidation and reduced nitrogen in the form of ammonium sulphate is liberated.}$ $\text{Organic compound} + \text{H}_2\text{SO}_4 \rightarrow [\text{digest}] \text{Cu}_2 + (\text{NH}_4)_2\text{SO}_4$ $\text{2. Distillation: The distillation of the solution now takes place and a small quantity of sodium hydroxide is added to convert the ammonium salt to ammonia.}$ $(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \xrightarrow{\Delta} \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{NH}_3$ $\text{3. Titration: The amount of ammonia or the amount of nitrogen present in the sample is then determined by back titration. As the ammonia dissolves in the acid trapping solution some HCl is neutralized. The acid that is left behind can be back titrated with a standard solution of a base such as NaOH or other bases.}$ $\text{B(OH)}_2 + \text{H}_2\text{O} + \text{Na}_2\text{CO}_3 \rightarrow \text{NaHCO}_3 + \text{CO}_2 + \text{H}_2\text{O}$ $\text{Kjeldahl's method is based on the fact that nitrogen of an organic compound is quantitatively converted to } (\text{NH}_4)_2\text{SO}_4 \text{ on heating with } \text{H}_2\text{SO}_4 \text{ (conc.). The } (\text{NH}_4)_2\text{SO}_4 \text{ is then treated with KOH to liberate } \text{NH}_3 \text{ which is absorbed } \text{H}_2\text{SO}_4 \text{ (conc.) to obtain \% of N.}$ $\text{So, Thus in Kjeldahl's method, nitrogen present is estimated as } \text{NH}_3\text{.}$

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{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}