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Current Question (ID: 7359)

Question:

$\text{1cc N}_2\text{O at STP contains:}$

Options:

1. $\frac{1.32}{224} \times 10^{23} \text{ electrons}$
2. $\frac{6.02}{22400} \times 10^{23} \text{ molecules}$
3. $\frac{1.8}{224} \times 10^{22} \text{ atoms}$
4. $\text{All of the above}$

Solution:

$\text{Hint: At STP, 1 mol gas volume is 22.4 L}$\n\n$\text{Step 1:}$\n\n$\text{As we know}$\n\n$\text{22400 cc of N}_2\text{O contain } 6.02 \times 10^{23} \text{ molecules}$\n\n$\text{Therefore 1 cc of N}_2\text{O contain } \frac{6.02 \times 10^{23}}{22400} \text{ molecules}$\n\n$\text{Since N}_2\text{O molecule there are 3 atoms}$\n\n$\text{Therefore 1 cc of N}_2\text{O} = \frac{3 \times 6.02 \times 10^{23}}{22400} \text{ atoms} = \frac{1.8 \times 10^{22}}{224} \text{ atoms}$\n\n$\text{Step 2:}$\n\n$\text{Number of electrons in molecule of N}_2\text{O} = 7 + 7 + 8 = 22$\n\n$\text{Hence number of electrons} = \frac{6.02 \times 10^{23}}{22400} \times 22 \text{ electrons} = \frac{1.32 \times 10^{23}}{224} \text{ electrons}$

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
}