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Current Question (ID: 21512)

Question:
$\text{For first-order kinetic rate constant } 2.011 \times 10^{-3} \text{ sec}^{-1}. \text{ The time taken for the decomposition of the substance from } 7 \text{ g to } 2 \text{ g will be:}$ $\text{(Use log } 7 = 0.845 \text{ and log } 2 = 0.301)$
Options:
  • 1. $647 \text{ sec}$
  • 2. $598 \text{ sec}$
  • 3. $623 \text{ sec}$
  • 4. $604 \text{ sec}$
Solution:
$\text{A} \rightarrow \text{products}$ $\text{Initial moles of A} = \frac{7}{M} \left( M \text{ is molar mass of A} \right)$ $\text{Final moles of A} = \frac{2}{M}$ $\text{Rate constant } k = 2.011 \times 10^{-3} \text{ s}^{-1}$ $\text{For a first order reaction}$ $t = \frac{2.303}{k} \log \frac{7}{2}$ $= \frac{2.303}{2.011} \times 10^{3} [0.845 - 0.301]$ $= 623 \text{ sec}$

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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
}