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Current Question (ID: 17953)

Question:

$\text{The density of KCl is } 1.9893 \text{ g cm}^{-3} \text{ and the length of a side unit cell is } 6.29082 \text{ \AA as determined by X-ray diffraction. The value of Avogadro's number calculated from this data is:}$

Options:

1. $6.017 \times 10^{23}$
2. $6.023 \times 10^{22}$
3. $7.03 \times 10^{23}$
4. $6.01 \times 10^{19}$

Solution:

$\text{HINT: } d(\text{density}) = \frac{Z \times M}{N_A \times V}$ $\text{STEP 1: } d = \frac{Z \cdot M}{N \cdot V}$ $Z = 4$ $d = 1.9893$ $\text{STEP 2: } a = 6.29082 \times 10^{-10}$ $V = a^3 = (6.29082 \times 10^{-10})^3$ $\text{STEP 3: } N = \frac{Z \cdot M}{d \cdot V}$ $= \frac{4 \times (39 + 35.5)}{1.989 \times (6.3 \times 10^{-10})^3}$ $= 6.017 \times 10^{23}$

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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
}