Import Question JSON

Current Question (ID: 9183)

Question:

$\text{The number of crystal water in Gypsum, Plaster of Paris and Epsom salt respectively are-}$

Options:

1. $2, 0.5, 7$
2. $7, 2, 1$
3. $7, 0.5, 2$
4. $3, 4, 2$

Solution:

$\text{HINT: Water of crystallization or water of hydration are water molecules that are present inside crystals.}$ $\text{Explanation:}$ $\text{The formula of Gypsum, Plaster of Paris and Epsom salt are:}$ $\text{CaSO}_4 \cdot \text{2H}_2\text{O}, \text{CaSO}_4 \cdot \text{0.5H}_2\text{O} \text{ and } \text{MgSO}_4 \cdot \text{7H}_2\text{O}$ $\text{So, the water of crystallization or number of crystal water in Gypsum, Plaster of Paris and Epsom salt respectively are } 2, 0.5 \text{ and } 7.$ $\text{1.) Gypsum is widely mined and is used as a fertilizer and as the main constituent in many forms of plaster, blackboard/sidewalk chalk, and drywall.}$ $\text{2.) Plaster of Paris is a building material that is used as a protective coating on walls and ceilings.}$ $\text{It is also used as a moulding and casting agent for decorative elements. It is used to give aesthetic finishing touches to the buildings.}$ $\text{3.) One of the most common uses for Epsom salt is to treat body aches.}$ $\text{The magnesium and other compounds are absorbed into your skin and work to relieve aches and pains caused from tension and inflammation.}$ $\text{Epsom salt draws toxins from your body to relieve swelling, sprains and bruises.}$

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
}