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Current Question (ID: 21369)

Question:

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

Options:

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

Solution:

$\text{Hint: 1 mole of electron charged is 1F}$ $\text{Step 1:}$ $\text{The cathode reaction is as follows:}$ $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$ $1\text{F is defined as a charge of 1 mole of electrons.}$ $\text{Hence, for } 27 \, \text{g of Al, a } 3\text{F charge is required.}$ $1\text{g of Al charge required} = \frac{3F}{27}$ $\text{For } 5.12 \, \text{kg of aluminum, charge required is} = \frac{3F}{27} \times 5.12 \times 10^3$ $= \frac{3 \times 96500}{27} \times 5.12 \times 10^3$ $= 5.49 \times 10^7 \, \text{C of electricity}$

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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
}