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Current Question (ID: 10976)

Question:
$\text{During estimation of nitrogen present in an organic compound, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of 1 M H}_2\text{SO}_4\text{. The percentage of nitrogen in the compound is-}$
Options:
  • 1. $46.0\%$
  • 2. $51.0\%$
  • 3. $56.0\%$ (Correct)
  • 4. $49.0\%$
Solution:
$\text{Hint: Percentage of nitrogen} = \frac{1.4 \times M \times 2 \left[V - \frac{V_1}{2}\right]}{m}$ $\text{Step 1:}$ $\text{1 M of 10 mL H}_2\text{SO}_4 = \text{1 M of 20 mL NH}_3$ $\text{1000 mL of 1M ammonia contains 14 g nitrogen}$ $\text{20 mL of 1M ammonia contains} = \frac{14}{1000} \times 20 = 0.28\text{g nitrogen}$ $\text{Step 2:}$ $\text{Percentage of nitrogen} = \frac{14}{1000} \times 20 \times \frac{100}{0.5} = 56.0\%$

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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
}