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Current Question (ID: 8360)

Question:
\text{Given the Gibbs free energy change, } \Delta G^{\circ} = +63.3 \text{ kJ, for the following reaction,} \text{Ag}_2\text{CO}_3\text{(s)} \rightarrow 2\text{Ag}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} K_{\text{sp}} \text{ of Ag}_2\text{CO}_3\text{(s)} \text{ in water at } 25°\text{C is:} (R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})
Options:
  • 1. $3.2 \times 10^{26}$
  • 2. $8.0 \times 10^{-12}$
  • 3. $2.9 \times 10^{-3}$
  • 4. $7.9 \times 10^{-2}$
Solution:
$\text{Hint: } \Delta G^o = -2.303 RT \log K_{\text{sp}}$\n\n$\text{Solution:}$\n\n$\text{STEP 1: } \Delta G^o = -2.303 RT \log K_{\text{sp}}$\n\n$63.3 \times 1000 \text{ J} = -2.303 (8.314)(298) \log K_{\text{sp}}$\n\n$\frac{-63.3 \times 1000}{2.303 (8.314)(298)} = \log K_{\text{sp}}$\n\n$-11.09 = \log K_{\text{sp}}$\n\n$\text{STEP 2: } K_{\text{sp}} = \text{antilog}(-11.09)$\n\n$\text{or } K_{\text{sp}} = 8.0 \times 10^{-12}$\n\n$\text{Thus, option 2 is the correct answer.}$

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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
}