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Current Question (ID: 20855)

Question:
$2.8 \times 10^{-3} \text{ mol of CO}_2 \text{ is left after removing } 10^{21} \text{ molecules from its } x \text{ mg sample. The mass of CO}_2 \text{ taken initially is:}$ $\text{(Given: } N_A = 6.02 \times 10^{23} \text{ mol}^{-1})$
Options:
  • 1. $196.2 \text{ mg}$
  • 2. $98.3 \text{ mg}$
  • 3. $150.4 \text{ mg}$
  • 4. $48.2 \text{ mg}$
Solution:
$(\text{moles})_{\text{initial}} = \frac{x \times 10^{-3}}{44}$ $\frac{10^{21}}{6.02 \times 10^{23}}$ $(\text{moles})_{\text{removed}} = \frac{10^{21}}{6.02 \times 10^{23}}$ $(\text{moles})_{\text{left}} = (\text{moles})_{\text{initial}} - (\text{moles})_{\text{removed}}$ $2.8 \times 10^{-3} = \frac{x \times 10^{-3}}{44} - \frac{10^{21}}{6.02 \times 10^{23}}$ $\Rightarrow x = 196.2 \text{ mg}$

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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
}