Import Question JSON

Current Question (ID: 10555)

Question:

$\text{Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.}$ $\text{Column I}$ $\text{A. Pb}_3\text{O}_4$ $\text{B. N}_2\text{O}$ $\text{C. Mn}_2\text{O}_7$ $\text{D. Bi}_2\text{O}_3$ $\text{Column II}$ $\text{1. Neutral oxide}$ $\text{2. Acidic oxide}$ $\text{3. Basic oxide}$ $\text{4. Mixed oxide}$

Options:

1. $\text{A-1, B-2, C-3, D-4}$
2. $\text{A-4, B-1, C-2, D-3}$ (Correct)
3. $\text{A-3, B-2, C-4, D-1}$
4. $\text{A-4, B-3, C-1, D-2}$

Solution:

$\text{HINT: Mn}_2\text{O}_7 \text{ is an acidic oxide.}$ $\text{Explanation:}$ $\text{Non metals react with oxygen to form acidic oxides and metals form basic oxides.}$ $\text{A mixed oxide is a somewhat informal name for an oxide that contains cations of more than one chemical element or cations of a single element in several states of oxidation.}$ $\text{Formulas of the compound and Type of oxide:}$ $\text{A. Pb}_3\text{O}_4 \text{ (PbO·Pb}_2\text{O}_3\text{) - Mixed oxide}$ $\text{B. N}_2\text{O} \text{ - Neutral oxide}$ $\text{C. Mn}_2\text{O}_7 \text{ - Acidic oxide}$ $\text{D. Bi}_2\text{O}_3 \text{ - Basic oxide}$ $\text{Mn}_2\text{O}_7 \text{ on dissolution in water produces acidic solution.}$ $\text{Bi}_2\text{O}_3 \text{ on dissolution in water produces basic solution.}$

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
}