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Current Question (ID: 7390)

Question:

$\text{On complete decomposition, the volume of } \text{CO}_2 \text{ released at STP on heating } 9.85 \text{ g of } \text{BaCO}_3 \text{ (atomic mass, Ba = 137) will be:}$

Options:

1. $1.12 \text{ L}$
2. $4.84 \text{ L}$
3. $2.12 \text{ L}$
4. $2.06 \text{ L}$

Solution:

$\text{Hint: At STP 1 mole is equal to 22.4 L}$\n\n$\text{Step 1:}$\n\n$\text{The reaction is as follows:}$\n\n$\text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2$\n\n$\text{When 197 grams of } \text{BaCO}_3 \text{ is decomposed it forms 22.4 liters of } \text{CO}_2$\n\n$\text{Number of moles} = \frac{9.85}{197} = 0.05$\n\n$\text{Step 2:}$\n\n$\text{1 mole produces 22.4 L}$\n\n$\text{so 0.05 mole produces } 0.05 \times 22.4 = 1.12 \text{ L}$\n\n$\text{Hence, the volume of } \text{CO}_2 \text{ released at STP on heating 9.85 g of } \text{BaCO}_3 \text{ is 1.12 L.}$

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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
}