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Current Question (ID: 18225)

Question:

$\text{A 100.0 mL dilute solution of } \text{Ag}^+ \text{ is electrolyzed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. The initial } [\text{Ag}^+] \text{ is:}$

Options:

1. $2.32 \times 10^{-1} \text{ M}$
2. $2.32 \times 10^{-4} \text{ M}$
3. $2.32 \times 10^{-5} \text{ M}$
4. $1.17 \times 10^{-4} \text{ M}$

Solution:

$\text{HINT: Faraday's first law of electrolysis.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Calculate the amount of Ag deposited}$ $\text{Ag}^+ + n\text{e} \rightarrow \text{Ag}$ $C = I \cdot t$ $= 1.25 \times 10^{-3} \times 15 \times 60 = 1.125 \text{ C}$ $96500 \text{ C} \xrightarrow{108 \text{ g Ag}}$ $1.125 = \frac{108}{96500} \times 1.125$ $\text{The amount of Ag deposited} = 1.26 \times 10^{-3} \text{ g}$ $\text{Step 2:}$ $\text{Calculate the concentration of } [\text{Ag}^+] \text{ ion}$ $\text{Molarity} = \frac{\text{amount of substance}}{\text{molar weight}} \times \frac{1000}{\text{volume of solution}}$ $[\text{Ag}^+] = \frac{1.26 \times 10^{-3}}{108} \times \frac{1000}{100} = 1.17 \times 10^{-4} \text{ M}$

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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
}