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Current Question (ID: 19863)

Question:

$\text{Number of molecules in a volume of } 4 \text{ cm}^3 \text{ of a perfect monoatomic gas at some temperature } T \text{ and at a pressure of } 2 \text{ cm of mercury is close to } ?$ $\text{(Given, mean kinetic energy of a molecule (at } T) \text{ is } 4 \times 10^{-14} \text{ erg, } g = 980 \text{ cm/s}^2, \text{ density of mercury } = 13.6 \text{ g/cm}^3 )$

Options:

1. $5.8 \times 10^{18}$
2. $5.8 \times 10^{16}$
3. $4.0 \times 10^{18}$
4. $4.0 \times 10^{16}$

Solution:

$\text{Hint: } PV = nRT$ $n = \frac{PV}{RT}, \quad \frac{3}{2} kT = 4 \times 10^{-14}$ $N = \frac{PV}{RT} \times N_a$ $= \frac{2 \times 13.6 \times 980 \times 4}{\frac{8}{3} \times 10^{-14}} = 3.99 \times 10^{18}$

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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
}