Import Question JSON

Current Question (ID: 21233)

Question:

$\text{In an estimation of bromine by the Carius method, } 1.6 \text{ g of an organic compound gave } 1.88 \text{ g of AgBr. The mass percentage of bromine in the compound is:}$ $[\text{Atomic mass, Ag}=108, \text{ Br} = 80 \text{ g mol}^{-1}]$

Options:

1. $50 \%$
2. $55.5 \%$
3. $60 \%$
4. $70 \%$

Solution:

$\text{The formula of percentage of halogen is as follows:}$ $\% \text{ of halogen} = \frac{\text{atomic mass of } X \times m_1 \times 100}{\text{molecular mass of AgX} \times m}$ $m_1 = \text{mass of AgX is formed}$ $m = \text{mass of organic compound}$ $\text{Calculate the } \% \text{ of halogen is as follows:}$ $\% \text{ of halogen} = \frac{80 \times 1.88 \times 100}{188 \times 1.6}$ $= 50 \%$

Import JSON File

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
}