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Current Question (ID: 21228)

Question:

$\text{In the Carius method of estimation of halogen, } 0.172 \text{g of an organic compound showed the presence of } 0.08 \text{g of bromine. The correct structure of the compound among the following is:}$

Options:

1. $\begin{array}{c} \text{NH}_2 \\ \text{Br} \\ \text{(benzene ring)} \end{array}$
2. $\text{H}_3\text{C} - \text{CH}_2 - \text{Br}$
3. $\begin{array}{c} \text{NH}_2 \\ \text{Br} \\ \text{(benzene ring)} \end{array}$
4. $\text{H}_3\text{C} - \text{Br}$

Solution:

$\text{Hint: The amount of Br in the organic compound is fixed}$ $\text{Step 1:}$ $\text{First, calculate the mass percentage of bromine in the given organic compound as follows:}$ $\text{In Carius method}$ $\text{mass of organic compound} = 0.172 \text{ gm}$ $\text{mass of Bromine} = 0.08 \text{ gm}$ $\text{Hence } \% \text{ of Bromine} = \frac{0.08}{0.172} \times 100 = 46.51\%$ $\text{The organic compound contains, } 46.51\% \text{ Br}$ $\text{Step 2:}$ $\text{Calculate the mass percentage of Br in every compound given in the options}$ $1.$ $\text{The mass percentage of Br in } \text{C}_6\text{H}_6\text{NBr} = \frac{80}{172} \times 100; = 46.51\%$ $2.$ $\text{CH}_3\text{CH}_2\text{Br} \left( \frac{\text{C}_2\text{H}_5\text{Br}}{\text{C}_2\text{H}_5\text{Br}} \right) \left[ \% \text{ Br} = \frac{80}{109} \times 100 \right] = 73.33\%$ $3.$ $\text{The mass percentage of Br in } \text{C}_6\text{H}_5\text{NBr}_2 = \frac{2 \times 80}{251} \times 100; = 63.74 \%$ $4.$ $\text{The mass percentage of Br in } \text{CH}_3\text{Br} = \frac{80}{95} \times 100; = 84.21 \%$ $\text{Hence, option first is the correct answer.}$

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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
}