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Current Question (ID: 18234)

Question:

$\text{The weight of silver (atomic weight = 108) displaced by a quantity of electricity which displaces 5600 mL of } \text{O}_2 \text{ at STP will be:}$

Options:

1. $5.4 \text{ g}$
2. $10.8 \text{ g}$
3. $54.0 \text{ g}$
4. $108.0 \text{ g}$

Solution:

$\text{HINT: Use Faraday's Second law of Electrolysis.}$ $\text{STEP 1:}$ $\text{Since, 22400 mL volume is occupied by 1 mole of } \text{O}_2 \text{ at STP.}$ $\text{Thus, } 5600 \text{ mL } \text{O}_2 \text{ means } = \frac{5600}{22400} \text{ mol } \text{O}_2 = \frac{1}{4} \text{ mol } \text{O}_2$ $\therefore \text{Weight of } \text{O}_2 = \frac{1}{4} \times 32 = 8 \text{ g}$ $\text{STEP 2:}$ $\text{According to problem,}$ $\text{Equivalents of Ag} = \text{Equivalents of } \text{O}_2$ $= \frac{\text{Weight}}{\text{Ag Equivalent weight}} = \frac{W_{\text{O}_2}}{\text{Equivalent weight of } \text{O}_2}$ $= \frac{W_{\text{O}_2}}{M_{\text{O}_2}/\text{valence}} \text{ factor}$ $\frac{W_{\text{Ag}}}{108} \times 1 = \frac{8}{32} \times 4 \quad [\because \text{ } 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-]$ $\therefore W_{\text{Ag}} = 108 \text{ g}$

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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
}