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Current Question (ID: 21388)

Question:
$\text{Electrolysis of } \text{Fe}_2(\text{SO}_4)_3 \text{ solution is carried out for 'x' minutes with a current of } 1.5 \text{ A, depositing } 0.3482 \text{ g of Fe. What is the value of } x \text{ to the nearest integer?}$ $\text{[Given: } 1 \text{ F } = 96500 \text{ C mol}^{-1}]$ $\text{Atomic mass of Fe } = 56 \text{ g mol}^{-1}$
Options:
  • 1. $20$
  • 2. $25$
  • 3. $18$
  • 4. $17$
Solution:
$\text{Fe}^{3+} + 3\text{e}^- \rightarrow \text{Fe}$ $\text{Moles of Fe deposited} = \frac{0.3482}{56} = 6.2 \times 10^{-3}$ $\text{For 1 mole Fe, charge required is 3 F}$ $\text{For } 6.2 \times 10^{-3} \text{ mole Fe, charge required is } 3 \times 6.2 \times 10^{-3} \text{ F}$ $\text{Since, charge required} = 18.6 \times 10^{-3} \times 96500 \text{C}$ $= 1794.9 \text{ C}$ $\text{And,}$ $1.5 \times t = 1794.9$ $t = \frac{1794.9}{1.5 \times 60} \text{ min}$ $t \simeq 20 \text{ min}$

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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
}