Import Question JSON

Current Question (ID: 21241)

Question:

$\text{An organic compound 'A' is oxidized with } \text{Na}_2\text{O}_2 \text{ followed by boiling with } \text{HNO}_3. \text{ The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on the above observation, the element present in the given compound is:}$

Options:

1. $\text{Fluorine}$
2. $\text{Phosphorus}$
3. $\text{Nitrogen}$
4. $\text{Sulphur}$

Solution:

$\text{Hint: A yellow colouration with ammonium molybdate indicates the presence of phosphorus.}$ $\text{Explanation:}$ $\text{A yellow coloration or precipitate when resultant solution reacts with ammonium molybdate indicates the presence of phosphorus.}$ $\text{The compound is first heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate.}$ $\text{The reactions are as follows:}$ $\text{Na}_3\text{PO}_4 + 3\text{HNO}_3 \rightarrow \text{H}_3\text{PO}_4 + 3\text{NaNO}_3$ $\text{H}_3\text{PO}_4 + 12(\text{NH}_4)_2\text{MoO}_4 + 21\text{HNO}_3 \rightarrow$ $(\text{NH}_4)_3\text{PO}_4\cdot12\text{MoO}_3 + 21\text{NH}_4\text{NO}_3 + 12\text{H}_2\text{O}$ $\text{Hence, option 2 is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "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
}