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Current Question (ID: 10983)

Question:

$0.1688 \text{ g organic compound when analyzed by the Dumas method yields } 31.7 \text{ mL of moist nitrogen measured at } 14°\text{C, and } 758 \text{ mm mercury pressure. The \% of nitrogen in the organic compound (Aqueous tension at } 14°\text{C} = 12 \text{ mm) is-}$

Options:

1. $30.9\%$
2. $10\%$
3. $40\%$
4. $21.9\%$ (Correct)

Solution:

$\text{Hint: Percentage of nitrogen} = \left(\frac{28}{22.4} \times \frac{V_0}{W}\right) \times 100$ $\text{Step 1:}$ $\text{Weight of Organic compound = 0.168 g}$ $\text{Volume of moist nitrogen } (V_1) = 31.7 \text{ mL} = 31.7 \times 10^{-3} \text{ L}$ $\text{Temperature } (T_1) = 14°\text{C} = 287 \text{ K}$ $\text{Pressure of Moist nitrogen (P) = 758 mm Hg}$ $\text{Aqueous tension at } 14°\text{C} = 12 \text{ mm of Hg}$ $\therefore \text{Pressure of dry nitrogen} = (P-P^0) = 758-12 = 746 \text{ mm of Hg}$ $\frac{P_1V_1}{T_1} = \frac{P_0V_0}{T_0}$ $\therefore V_0 = \frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760}$ $V_0 = 29.58 \times 10^{-3} \text{ L}$ $\text{Step 2:}$ $\text{Percentage of nitrogen} = \left(\frac{28}{22.4} \times \frac{V_0}{W}\right) \times 100$ $= \frac{28}{22.4} \times \frac{29.58 \times 10^{-3}}{0.1688} \times 100$ $= 21.90\%$

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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
}