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Current Question (ID: 21244)

Question:

$\text{In the Carius method for estimation of halogens, 250 mg of an}$ $\text{organic compound gave 141 mg of AgBr. The percentage of bromine}$ $\text{in the compound is:}$ $\text{(Atomic masses: Ag = 108, Br = 80)}$

Options:

1. $24\%$
2. $36\%$
3. $48\%$
4. $60\%$

Solution:

$\text{Hint: Use the formula, Percentage of halogen = }$ $\frac{\text{atomic mass of } X \times m_1 \times 100}{\text{molecular mass of } \text{AgX} \times m}$ $\text{The formula of percentage of halogen is as follows:}$ $\% \text{ of halogen} = \frac{\text{atomic mass of } X \times m_1 \times 100}{\text{molecular mass of } \text{AgX} \times m}$ $m_1 = \text{mass of AgX is formed}$ $m = \text{mass of organic compound}$ $\text{Calculate the } \% \text{ of halogen is as follows:}$ $\% \text{ of halogen} = \frac{80 \times 141 \times 100}{188 \times 250}$ $= \frac{11280}{47000}$ $= 24\%$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}