Import Question JSON

Current Question (ID: 9209)

Question:

$\text{CaCl}_2 \text{ is preferred over NaCl for clearing ice on roads particularly in very cold countries. This is because:}$

Options:

1. $\text{CaCl}_2 \text{ is less soluble in } \text{H}_2\text{O} \text{ than NaCl.}$
2. $\text{CaCl}_2 \text{ is hygroscopic but NaCl is not.}$
3. \text{Eutectic mixture of } \text{CaCl}_2/\text{H}_2\text{O} \text{ freezes at } -55°\text{C} \text{ while that of } \text{NaCl}/\text{H}_2\text{O} \text{ freezes at } -18°\text{C}.
4. $\text{NaCl makes the road slippery but } \text{CaCl}_2 \text{ does not.}$

Solution:

$\text{HINT: Due to much lower freezing point of eutectic mixture of } \text{CaCl}_2/\text{H}_2\text{O}.$ $\text{Explanation:}$ $\text{1.) } \text{CaCl}_2 \text{ is more soluble in water than NaCl.}$ $\text{2.) Substance that can absorb moisture from the air is called hygroscopic. Both } \text{CaCl}_2 \text{ and NaCl are hygroscopic in nature.}$ $\text{3.) Eutectic mixture of } \text{CaCl}_2/\text{H}_2\text{O} \text{ freezes at } -55°\text{C} \text{ while that of NaCl}/\text{H}_2\text{O} \text{ freezes at } -18°\text{C}, \text{ that is why } \text{CaCl}_2 \text{ is preferred over NaCl for clearing ice on roads particularly in very cold countries.}$ $\text{4.) Both NaCl and } \text{CaCl}_2 \text{ can make the roads slippery.}$

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
}