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Current Question (ID: 8154)

Question:

The volume occupied by 8.8 g of $\text{CO}_2$ at 31.1$^\circ\text{C}$ and 1 bar pressure is: \n\n($\text{R} = 0.083 \text{ bar L K}^{-1} \text{mol}^{-1}$)

Options:

1. $4.05 \text{ L}$
2. $3.05 \text{ L}$
3. $6.05 \text{ L}$
4. $5.05 \text{ L}$

Solution:

Hint: Use Ideal gas equation\n\nStep 1: Write down the given data\n$\text{m} = 8.8 \text{ g}$\n$\text{R} = 0.083 \text{ bar L K}^{-1} \text{mol}^{-1}$\n$\text{T} = 31.1^\circ\text{C} = 304.1 \text{ K}$\n$\text{M} = 44 \text{ g/mol}$\n$\text{p} = 1 \text{ bar}$\n\nStep 2: Use Ideal gas equation to calculate volume\nIt is known that,\n$\text{pV} = \frac{\text{m}}{\text{M}} \text{RT}$\n$\Rightarrow \text{V} = \frac{\text{mRT}}{\text{Mp}}$\n\nThus, volume ($\text{V}$) $= \frac{8.8 \times 0.083 \times 304.1}{44 \times 1}$\n$= 5.04806 \text{ L}$\n$= 5.05 \text{ L}$

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
}