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Current Question (ID: 18342)

Question:
$\text{If } 60\% \text{ of a first-order reaction is completed in } 60 \text{ minutes, } 50\% \text{ of the same reaction takes approximately:}$ $\text{(log4 = 0.60, log5 = 0.69)}$
Options:
  • 1. $55 \text{ min}$
  • 2. $45 \text{ min}$
  • 3. $60 \text{ min}$
  • 4. $30 \text{ min}$
Solution:
$\text{HINT: } k = \frac{2.303}{t} \log \left( \frac{a}{a-x} \right)$ $k = \frac{2.303}{t_1} \log \frac{a_1}{a_1-x_1}$ $k = \frac{2.303}{t_2} \log \frac{a_2}{a_2-x_2}$ $x_1 = 60 \ a_1, t_1 = 60$ $x_2 = 50 \ a_2, t_2 = ?$ $\frac{2.303}{60} \log \frac{100 \ a_1}{40 \ a_1} = \frac{2.303}{t_2} \log \frac{100 \ a_2}{50 \ a_2}$ $\frac{1}{60} \log \frac{100}{40} = \frac{1}{t_2} \log \frac{100}{50}$ $t_2 = \frac{60 \log 100/50}{\log 100/40}$ $= \frac{60 \log 10 - \log 5}{(\log 10 - \log 4)}$ $= \frac{60 (1 - 0.69)}{0.40}$ $= \frac{60 \times 0.31}{0.40}$ $= 1.5 \times 31 = 46.5 \approx 45 \text{ min}$

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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
}