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Current Question (ID: 7383)

Question:
$\text{On complete combustion, 44 g of a sample of a compound gives 88 g CO}_2 \text{ and 36 g of H}_2\text{O. The molecular formula of the compound may be:}$
Options:
  • 1. $\text{C}_4\text{H}_6$
  • 2. $\text{C}_2\text{H}_6\text{O}$
  • 3. $\text{C}_2\text{H}_4\text{O}$
  • 4. $\text{C}_3\text{H}_6\text{O}$
Solution:
$\text{Hint: Empirical formula concept}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Calculate the moles of carbon and hydrogen as follows:}$ $\text{CO}_2 = \frac{88}{44} = 2 \text{ moles of CO}_2 = 2 \text{ mole of C}$ $\text{H}_2\text{O} = \frac{36}{18} = 2 \text{ moles of H}_2\text{O} = 4 \text{ moles of H}$ $\text{Step 2:}$ $\text{Calculate the empirical formula as follows:}$ $\text{Mass of C + Mass of H + Mass of O} = 44$ $\Rightarrow 24 + 4 + x = 44; x = 16$ $\therefore \text{mole of O} = 1 \text{ and molecular formula is C}_2\text{H}_4\text{O}$ $\text{Hence, option third is the correct answer.}$

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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
}