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Current Question (ID: 10958)

Question:

$\text{In Kjeldahl's method for estimation of nitrogen present in the soil sample, ammonia evolved from 0.75g of sample neutralized 10ml. of 1M H}_2\text{SO}_4\text{. The percentage of nitrogen in the soil is:}$

Options:

1. $37.33$ (Correct)
2. $45.85$
3. $25.75$
4. $43.13$

Solution:

$\text{Hint:}$ $\text{Step 1:}$ $\text{1 M of 10 mL H}_2\text{SO}_4 = \text{1M of 20 mL NH}_3$ $\text{1000 mL of 1M ammonia contains 14 g nitrogen}$ $\text{1 mL of 1M ammonia contains } \frac{14}{1000} \text{ g of nitrogen}$ $\text{20 mL of 1M ammonia contains } = \frac{14}{1000} \times 20 \text{ g of nitrogen}$ $= 0.28 \text{ g nitrogen}$ $\text{Step 2:}$ $\% \text{ of nitrogen} = \frac{\text{amount of nitrogen}}{\text{amount of organic compound}} \times 100$ $= \frac{0.28}{0.75} \times 100$ $= 37.3\%$

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
}