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Current Question (ID: 7397)

Question:

$\text{20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g of magnesium oxide.}$\n\n$\text{The percentage purity of magnesium carbonate in the sample is:}$\n\n$\text{(Atomic weight of Mg = 24)}$

Options:

1. $75$
2. $96$
3. $60$
4. $84$

Solution:

$\text{Hint: Percent purity} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$\n\n$\text{Step 1:}$\n\n$\text{To calculate the percentage purity of } \text{MgCO}_3\text{, we need to find the theoretical yield of MgO from pure } \text{MgCO}_3\text{.}$\n\n$\text{The decomposition reaction is:}$\n\n$\text{MgCO}_3\text{(s)} \rightarrow \text{MgO(s)} + \text{CO}_2\text{(g)}$\n\n$\text{Step 2: Calculate the molecular weights}$\n\n$\text{Molecular weight of } \text{MgCO}_3 = 24 + 12 + 3 \times 16 = 84 \text{ g/mol}$\n\n$\text{Molecular weight of MgO} = 24 + 16 = 40 \text{ g/mol}$\n\n$\text{Step 3: Calculate the theoretical amount of } \text{MgCO}_3 \text{ in the sample}$\n\n$\text{Number of moles of } \text{MgCO}_3 \text{ in the sample} = \frac{20.0 \text{ g}}{84 \text{ g/mol}} = 0.238 \text{ mol}$\n\n$\text{Step 4: Calculate the theoretical yield of MgO}$\n\n$\text{According to the stoichiometry of the reaction, 1 mole of } \text{MgCO}_3 \text{ gives 1 mole of MgO}$\n\n$\text{So, 0.238 moles of } \text{MgCO}_3 \text{ will give 0.238 moles of MgO}$\n\n$\text{Mass of MgO expected from pure } \text{MgCO}_3 = 0.238 \text{ mol} \times 40 \text{ g/mol} = 9.52 \text{ g}$\n\n$\text{Step 5: Calculate the percentage purity}$\n\n$\text{Actual yield of MgO} = 8.0 \text{ g (given)}$\n\n$\text{Percentage purity of } \text{MgCO}_3 = \frac{8.0 \text{ g}}{9.52 \text{ g}} \times 100\% = 84\%$\n\n$\text{Therefore, the percentage purity of magnesium carbonate in the sample is 84\%.}$

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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
}