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Current Question (ID: 7406)

Question:

$\text{10 g of a silver coin, when dissolved completely in excess of conc. } \text{HNO}_3 \text{ gives 8.5 g of silver nitrate.}$\n\n$\text{The percentage purity of the coin is:}$\n\n$\text{(Molar mass of silver is 108 g)}$

Options:

1. $25\%$
2. $54\%$
3. $67\%$
4. $100\%$

Solution:

$\text{Hint: Percentage Purity} = \frac{\text{actual amount of substance}}{\text{theoretical amount of substance}} \times 100$\n\n$\text{Step 1: Write the balanced chemical equation for the reaction of silver with concentrated nitric acid.}$\n\n$\text{Ag} + 2\text{HNO}_3 \text{ (conc.)} \rightarrow \text{AgNO}_3 + \text{NO}_2 + \text{H}_2\text{O}$\n\n$\text{Step 2: Calculate the molar mass of silver nitrate.}$\n\n$\text{Molar mass of } \text{AgNO}_3 = 108 + 14 + 3 \times 16 = 108 + 14 + 48 = 170 \text{ g/mol}$\n\n$\text{Step 3: Determine the relationship between silver and silver nitrate.}$\n\n$\text{From the balanced equation, 1 mole of Ag forms 1 mole of } \text{AgNO}_3$\n\n$\text{108 g of Ag forms 170 g of } \text{AgNO}_3$\n\n$\text{Step 4: Calculate the amount of silver that would yield 8.5 g of silver nitrate.}$\n\n$\text{Amount of Ag in 8.5 g of } \text{AgNO}_3 = \frac{108 \text{ g}}{170 \text{ g}} \times 8.5 \text{ g} = 5.4 \text{ g}$\n\n$\text{Step 5: Calculate the percentage purity of the silver coin.}$\n\n$\text{Percentage purity} = \frac{\text{Mass of Ag in the coin}}{\text{Total mass of the coin}} \times 100\%$\n\n$\text{Percentage purity} = \frac{5.4 \text{ g}}{10 \text{ g}} \times 100\% = 54\%$\n\n$\text{Therefore, the percentage purity of the silver coin is 54\%.}$

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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
}