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Current Question (ID: 21255)

Question:

$\text{The ammonia evolved from the treatment of } 0.30 \text{ g of an organic compound for the estimation of nitrogen was passed in } 100 \text{ mL of } 0.1 \text{ M sulphuric acid. The excess acid required } 20 \text{ mL of } 0.5 \text{ M sodium hydroxide solution for complete neutralization. The percentage of nitrogen in the organic compound is:}$

Options:

1. $46.6 \%$
2. $50.4 \%$
3. $42.8 \%$
4. $40.5 \%$

Solution:

$\text{Hint: } \% \text{ of } \text{N}_2 = \frac{1.4 \times \text{Meq of acid neutralised by } \text{NH}_3}{\text{weight of organic compound}}$ $\text{Step 1:}$ $\text{H}_2\text{SO}_4 \text{ is a dibasic acid.}$ $\text{Meq of } \text{H}_2\text{SO}_4 \text{ taken } = 2 \times 0.1 \times 100 = 20$ $\text{Meq of } \text{H}_2\text{SO}_4 \text{ neutralized by } \text{NaOH} = 0.5 \times 20 = 10$ $\text{Meq of } \text{H}_2\text{SO}_4 \text{ neutralized by } \text{NH}_3 = 20 - 10 = 10$ $\text{Step 2:}$ $\% \text{ of } \text{N}_2 = \frac{1.4 \times \text{Meq of acid neutralised by } \text{NH}_3}{\text{weight of organic compound}}$ $\% \text{ of nitrogen in urea } = \frac{1.4 \times 10}{0.3} = 46.6$

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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
}