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Current Question (ID: 8417)

Question:

$\text{For the reaction, } 3\text{O}_2\text{(g)} \rightleftharpoons 2\text{O}_3\text{(g)}, K_c = 2.0 \times 10^{-50} \text{ L/mol at } 25°\text{C. If the equilibrium concentration of O}_2 \text{ in air at } 25°\text{C is } 1.6 \times 10^{-2}, \text{ the concentration of O}_3 \text{ would be:}$

Options:

1. $2.86 \times 10^{28} \text{ M}$
2. $28.66 \times 10^{-28} \text{ M}$
3. $1.43 \times 10^{-28} \text{ M}$
4. $2.86 \times 10^{-28} \text{ M}$

Solution:

$\text{Hint: Use general information of } K_c$ $\text{Explanation:}$ $\text{STEP 1: Write down the given data}$ $K_c = 2.0 \times 10^{-50} \text{ and } [\text{O}_2] = 1.6 \times 10^{-2}$ $\text{STEP 2: Find out } [\text{O}_3] \text{ using general expression of } K_c$ $K_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}$ $\Rightarrow 2.0 \times 10^{-50} = \frac{[\text{O}_3]^2}{(1.6 \times 10^{-2})^3}$ $\Rightarrow [\text{O}_3]^2 = 8.19 \times 10^{-56}$ $\Rightarrow [\text{O}_3] = 2.86 \times 10^{-28}$ $\text{Hence, the concentration of O}_3 \text{ is } 2.86 \times 10^{-28} \text{ M}$

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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
}