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Current Question (ID: 7381)

Question:

$\text{The empirical formula and molecular mass of a compound are CH}_2\text{O and 180 g, respectively. The molecular formula of the compound is:}$

Options:

1. $\text{C}_9\text{H}_{18}\text{O}_9$
2. $\text{CH}_2\text{O}$
3. $\text{C}_6\text{H}_{12}\text{O}_6$
4. $\text{C}_2\text{H}_4\text{O}_2$

Solution:

$\text{Hint: n} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}$ $\text{Empirical formula mass}= \text{CH}_2\text{O}$ $\text{Formula mass is}$ $= 12 + 2 \times 1 + 16 = 30$ $\text{Molecular mass} = 180$ $\text{n} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}$ $= \frac{180}{30} = 6$ $\therefore \text{Molecular formula} = \text{n} \times \text{empirical formula}$ $= 6 \times \text{CH}_2\text{O}$ $= \text{C}_6\text{H}_{12}\text{O}_6$

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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
}