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Current Question (ID: 10984)

Question:

$\text{In Dumas method of estimation of nitrogen, 0.35 g of an organic compound gave 55 ml of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be:}$ $\text{(Aqueous tension at 300 K = 15 mm)}$

Options:

1. $16.45$ (Correct)
2. $27.45$
3. $44.45$
4. $35.45$

Solution:

$\text{Hint: } \% \text{ of N} = \frac{28}{22400} \times \frac{\text{volume of } N_2 \text{ at STP(in) ml}}{\text{weight of organic compound (in) g}} \times 100$ $\text{Step 1:}$ $\text{Calculate the volume of nitrogen at STP as follows:}$ $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \text{ (Here,) } P_1 = 715 - 15 = 700 \text{ mm}$ $\frac{700 \times 55}{300} = \frac{760 \times V_2}{273}$ $V_2 = 46 \text{ ml at STP}$ $\text{Step 2:}$ $\% \text{ of N} = \frac{28}{22400} \times \frac{\text{volume of } N_2 \text{ at STP in ml}}{\text{wt. of org. compound in g}} \times 100$ $= \frac{28}{22400} \times \frac{46}{0.35} \times 100 = 16.45$

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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
}