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Current Question (ID: 18338)

Question:
$\text{The half-life time for the decomposition of a substance dissolved in } \text{CCl}_4 \text{ is } 2.5 \text{ hours at } 30^\circ \text{C. The amount of substance that will be left after } 10 \text{ hours if the initial weight of the substance is } 160 \text{ gm is:}$
Options:
  • 1. $20 \text{ gm}$
  • 2. $30 \text{ gm}$
  • 3. $40 \text{ gm}$
  • 4. $10 \text{ gm}$
Solution:
$\text{HINT: Find number of Half-life of the reaction}$ $\text{Explanation: No. of half life periods} = \frac{\text{Total time}}{\text{Time of one half period}} = \frac{10}{2.5} = 4$ $\text{Now we know that the quantity left after 'n' half-life periods}$ $= \left(\frac{1}{2}\right)^n \times \text{Initial quantity}$ $= \left(\frac{1}{2}\right)^4 \times 160$ $= \frac{160}{2^4} = 10 \text{ gm}$

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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
}