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Current Question (ID: 21505)

Question:
$\text{The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, then the mass remaining undecayed after 24 hours is:}$ $1.\ 1.042\ \text{g}$ $2.\ 2.084\ \text{g}$ $3.\ 3.125\ \text{g}$ $4.\ 4.167\ \text{g}$
Options:
  • 1. $1.042\ \text{g}$
  • 2. $2.084\ \text{g}$
  • 3. $3.125\ \text{g}$
  • 4. $4.167\ \text{g}$
Solution:
$\text{Hint: Radioactive reaction follows first-order kinetics}$ $\text{Step 1:}$ $\text{Calculate the value of the rate constant as follows:}$ $t_{1/2} = \frac{0.693}{k}$ $k = \frac{0.693}{4} = 0.173\ \text{hr}^{-1}$ $\text{Step 2:}$ $\text{Calculate the mass of the isotope after 24 hours as follows:}$ $\text{Use the integrated equation for first-order as follows:}$ $k = \frac{2.303}{t} \log \frac{[A_0]}{[A]}$ $0.173 = \frac{2.303}{24} \log \frac{200}{[A]}$ $1.81 = \log \frac{200}{[A]}$ $\text{Take antilog both sides as follows:}$ $10^{1.81} = \frac{200}{[A]}$ $[A] = \frac{200}{64}$ $= 3.125\ \text{g}$ $\text{Hence, the most appropriate answer is option third}$

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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
}