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Current Question (ID: 8201)

Question:

The unit of van der Waals' constant 'b' is -

Options:

1. $\text{cm}^3 \text{mol}^{-1}$
2. $\text{litre mol}^{-1}$
3. $\text{m}^3 \text{mol}^{-1}$
4. All of the above

Solution:

Hint: b is the van der Waals constant.\n\nExplanation:\n\n"b" corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered.\n\nUnits of (a)\n\n$P = \frac{an^2}{V^2}$\n\n$a = \frac{PxV^2}{n^2} = \frac{\text{atm} \times \text{L}^2}{\text{mol}^2} = \text{atm L}^2 \text{mol}^{-2}$\n\nUnits of b\n\n$V = nb$\n\n$b = \frac{V}{n} = \frac{\text{L}}{\text{mol}} = \text{L mol}^{-1}$\n\nUnit of b = $\text{litre mol}^{-1} = \text{cm}^3 \text{mol}^{-1} = \text{m}^3 \text{mol}^{-1}$

Import JSON File

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
}