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Current Question (ID: 10977)

Question:

$\text{An organic compound contains 69% carbon, and 4.8% hydrogen, the remainder being oxygen. The masses of carbon dioxide, and water produced when 0.20 g of this substance is subjected to complete combustion would be respectively -}$

Options:

1. $0.506 \text{ g}, 0.0864 \text{ g}$ (Correct)
2. $0.906 \text{ g}, 0.0864 \text{ g}$
3. $0.0506 \text{ g}, 0.864 \text{ g}$
4. $0.0864 \text{ g}, 0.506 \text{ g}$

Solution:

$\text{Step 1:}$ $\text{Find the empirical formula of the given reactant and then predict the reaction of combustion. Use the % of the mass formula.}$ $\text{Percentage of carbon in organic compound = 69%}$ $\text{That is, 100 g of organic compound contains 69 g of carbon.}$ $0.2 \text{ g of the organic compound will contain} = \frac{69 \times 0.2}{100} = 0.138 \text{ of C}$ $\text{The molecular mass of carbon dioxide, } \text{CO}_2 = 44 \text{ g}$ $\text{That is, 12 g of carbon is contained in 44 g of } \text{CO}_2\text{.}$ $\text{Therefore, 0.138 g of carbon will be contained in } \frac{44 \times 0.138}{12} = 0.506 \text{ g of } \text{CO}_2$ $\text{Thus, 0.506 g of } \text{CO}_2 \text{ will be produced on complete combustion of 0.2 g of an organic compound.}$ $\text{Step 2:}$ $\text{The percentage of hydrogen in an organic compound is 4.8.}$ $\text{i.e., 100 g of organic compound contains 4.8 g of hydrogen.}$ $\text{Therefore, 0.2 g of the organic compound will contain } \frac{4.8 \times 0.2}{100} = 0.0096 \text{ g of H}$ $\text{It is known that the molecular mass of water } (\text{H}_2\text{O}) \text{ is 18 g.}$ $\text{Thus, 2 g of hydrogen is contained in 18 g of water.}$ $0.0096 \text{ g of hydrogen will be contained in water } \frac{18 \times 0.0096}{2} = 0.0864 \text{ g}$ $\text{Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic compound.}$

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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
}