Import Question JSON

Current Question (ID: 10638)

Question:

$\text{An amorphous solid "A" burns in the air to form a gas "B" which turns lime}$ $\text{water milky. The gas is also produced as a by-product during the roasting of}$ $\text{sulphide ore. This gas decolourises acidified aqueous KMnO}_4 \text{ solution and}$ $\text{reduces Fe}^{3+} \text{ to Fe}^{2+}\text{. The solid "A" and the gas "B" are:}$

Options:

1. $\text{'A' = S}_8\text{; 'B' = SO}_2$ (Correct)
2. $\text{'A' = S}_8\text{; 'B' = SO}_3$
3. $\text{'A' = S}_{10}\text{; 'B' = SO}_2$
4. $\text{Not predictable}$

Solution:

$\textbf{HINT:} \text{SO}_2 \text{ can decolourises acidified aq. KMnO}_4$$ $\textbf{Explanation:}$$ $\text{Since, the by-product of roasting of sulphide ore is SO}_2 \text{ so A is S}_8\text{, so 'A' = S}_8\text{:}$ $\text{'B'= SO}_2$$ $\textbf{Reactions}$$ $\text{(i) } \text{S}_8 + 8\text{O}_2 \xrightarrow{\Delta} 8\text{SO}_2$$ $\text{(ii) } \text{Ca(OH)}_2 + \text{SO}_2 \rightarrow \text{CaSO}_3 + \text{H}_2\text{O}$$ $\text{(iii) } 2\text{MnO}_4^- + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{Mn}^{2+}$$ $\text{(Violet)} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{(Colourless)}$$ $\text{(iv) } 2\text{Fe}^{3+} + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Fe}^{2+} + \text{SO}_4^{2-} + 4\text{H}^+$

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
}