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Current Question (ID: 19835)

Question:

$\text{The temperature of an open room of volume } 30 \text{ m}^3 \text{ increases from } 17^\circ\text{C to } 27^\circ\text{C due to the sunshine.}$ $\text{The atmospheric pressure in the room remains } 1 \times 10^5 \text{ Pa.}$ $\text{In } n_i \text{ and } n_f \text{ are the number of molecules in the room before and after heating, the } n_f-n_i \text{ will be:}$

Options:

1. $-1.61 \times 10^{23}$
2. $1.38 \times 10^{23}$
3. $2.5 \times 10^{25}$
4. $-2.5 \times 10^{25}$

Solution:

$\text{Hint: } PV = nRT$ $PV = nRT$ $n = \frac{PV}{RT}$ $T_i = 273 + 17$ $= 290 \text{ K}$ $T_f = 273 + 27$ $= 300 \text{ K}$ $n_f - n_i = \frac{10^5 \times 30}{8.314} \left[ \frac{1}{300} - \frac{1}{290} \right] \times 6.023 \times 10^{23}$ $= \frac{3 \times 10^6}{8.314} \left[ \frac{-10}{300 \times 290} \right] \times 6.023 \times 10^{23}$ $= \frac{-3 \times 10^{27} \times 6.023}{8.314 \times 3 \times 29}$ $= \frac{-6.023 \times 10^{27}}{8.314 \times 29}$ $= -0.025 \times 10^{27}$ $= -2.5 \times 10^{25}$

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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
}