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Current Question (ID: 17936)

Question:

$\text{Iron has a body centred cubic unit cell with the cell dimension of } 286.65 \text{ pm.}$ $\text{Density of iron is } 7.87 \text{ g cm}^{-3}. \text{ The value of Avogadro's number based on this data will be -}$ $\text{(Atomic mass of Fe = 56.0 u)}$

Options:

1. $6.04 \times 10^{23} \text{ mol}^{-1}$
2. $12.08 \times 10^{24} \text{ mol}^{-1}$
3. $6 \times 10^{22} \text{ mol}^{-1}$
4. $6 \times 10^{-23} \text{ mol}^{-1}$

Solution:

$\text{HINT: } d = \frac{Z \times M}{a^3 \times N_A},$ $\text{STEP 1:}$ $Z \text{ for bcc } = 2, \ a = 286.65 \text{ pm} = 286.65 \times 10^{-10} \text{ cm},$ $\text{Density } = 7.87 \text{ g cm}^{-3} \text{ and } M = 56 \text{ g mol}^{-1}$ $\text{STEP 2:}$ $N_A = \frac{Z \times M}{a^3 \times d} = \frac{2 \times 56}{(286.65 \times 10^{-10})^3 \times 7.87} \text{ g mol}^{-1}$ $= 6.043 \times 10^{23} \text{ mol}^{-1}$

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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
}