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Current Question (ID: 9195)

Question:
$\text{Gypsum on heating can change to :}$
Options:
  • 1. $\text{Orthorhombic form.}$
  • 2. $\text{Plaster of paris.}$
  • 3. $\text{Dead plaster.}$
  • 4. $\text{All of the above.}$
Solution:
$\text{Hint: Chemical name of Gypsum is calcium sulphate dihydrate (CaSO}_4 \cdot 2\text{H}_2\text{O}).$ $\text{Explanation:}$ $\text{Gypsum on heating first changes from monoclinic to orthorhombic form without loss of water at 120°C, it loses } \frac{3}{4}th \text{ of its water of crystallization and form plaster of Paris. The reaction is as follows:}$ $\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{120°\text{C}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$ $\text{Gypsum} \qquad \qquad \text{Plaster of Paris}$ $\text{While on heating at 200°C it changes to dead plaster of burnt plaster.}$

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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
}