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Current Question (ID: 18231)

Question:

$\text{Aluminium oxide may be electrolysed at } 1000 \ C \ \text{to furnish aluminium metal}$ $\text{(Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is:}$ $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ $\text{To prepare } 5.12 \ \text{kg of aluminium metal by this method, would require:}$

Options:

1. $5.49 \times 10^7 \ C \ \text{of electricity}$
2. $1.83 \times 10^7 \ C \ \text{of electricity}$
3. $5.49 \times 10^4 \ C \ \text{of electricity}$
4. $5.49 \times 10^9 \ C \ \text{of electricity}$

Solution:

$\text{The reaction is as follows:}$ $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ $1 \ \text{mole } \text{Al}^{3+} \ \text{generated 1 mole Al}$ $27 \ \text{g Al need } 3F \ \text{charge}$ $1 \ \text{g Al needs } \frac{3F}{27} \ \text{charge}$ $5.12 \ \text{kg needs } \frac{3F}{27} \times 5.12 \times \frac{1000}{1} \ \text{g1 kg}$ $= \frac{3 \times 96500}{27} \times 5.12 \times \frac{1000}{1} \ \text{g1 kg}$ $= 5.49 \times 10^7 \ C$ $\text{Hence, option 1 is the answer.}$

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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
}