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Current Question (ID: 21248)

Question:

$29.5 \text{ mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia were absorbed in } 20 \text{ mL of } 0.1 \text{ M HCl solution.}$ $\text{The excess of the acid required } 15 \text{ mL of } 0.1 \text{ M NaOH solution for complete neutralization.}$ $\text{The percentage of nitrogen in the compound is:}$

Options:

1. $29.5$
2. $23.7$
3. $47.4$
4. $59.0$

Solution:

$\text{Hint: Use the formula, that is, Percentage of nitrogen = } \frac{1.4 \times M \times (V - V_1)}{m}$ $\text{Step 1:}$ $\text{The formula of the percentage of nitrogen in an organic compound is given below when } \text{H}_2\text{SO}_4 \text{ is used in Kjeldahl's method.}$ $\text{Percentage of Nitrogen} = \frac{1.4 \times M \times 2 \left(V - \frac{V_1}{2}\right)}{m}$ $\text{But in the question, HCl is given hence, the formula will modify a little bit. The modified formula is as follows:}$ $\text{Percentage of nitrogen} = \frac{1.4 \times M \times (V - V_1)}{m}$ $\text{Here, M is the molarity of HCl, V is a volume of HCl and } V_1 \text{ is a volume of NaOH and m is the amount of organic compound in g.}$ $\text{Step 2:}$ $\text{Calculate the percentage of nitrogen as follows:}$ $\text{Percentage of nitrogen} = \frac{1.4 \times 0.1 \times (20 - 15)}{29.5 \times \frac{1}{1000}}$ $\text{Percentage of nitrogen} = 23.7 \%$ $\text{Hence, option second is the correct option.}$

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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
}