Import Question JSON

Current Question (ID: 21515)

Question:

$\text{A reaction follows 1}^{\text{st}} \text{ order kinetics with rate constant } (k) = 20 \text{ min}^{-1}. \text{ Calculate the time required to reach the concentration to } 1/32 \text{ times of initial concentration.}$

Options:

1. $0.17325 \text{ min}$
2. $1.7325 \text{ min}$
3. $17.325 \text{ min}$
4. $173.25 \text{ min}$

Solution:

$K = 20 \text{ min}^{-1}$ $t_{1/2} = \frac{0.693}{K} = \frac{0.693}{20} \text{ min}$ $C = \frac{C_0}{(C)^n} = \frac{C_0}{32}$ $C = \text{Concentration at time } t$ $C_0 = \text{Initial concentration}$ $n = \text{no of half life's}$ $n = 5$ $t = 5 \times t_{1/2}$ $= 5 \times \frac{0.693}{20} = 0.17325 \text{ min}$

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
}