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Current Question (ID: 17945)

Question:

$\text{Lithium has a bcc structure. Its density is } 530 \text{ kg m}^{-3} \text{ and its atomic mass is } 6.94 \text{ g mol}^{-1}. \text{ The edge length of a unit cell of lithium metal is: } (N_A = 6.02 \times 10^{23} \text{ mol}^{-1})$

Options:

1. $352 \text{ pm}$
2. $527 \text{ pm}$
3. $264 \text{ pm}$
4. $154 \text{ pm}$

Solution:

$\text{HINT: } d = \frac{ZM}{a^3N_A}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{The density } d = 530 \text{ kg/m}^3 = 0.53 \text{ g/cm}^3$ $(\therefore 1 \text{ kg/m}^3 = 10^{-3} \text{ g/cm}^3)$ $\text{Atomic mass } M = 6.94 \text{ g/mol}$ $\text{For bcc unit cell, the number of atoms per unit cell } Z = 2$ $\text{STEP 2:}$ $\text{Density } d = \frac{ZM}{a^3N_A}$ $0.530 = \frac{2 \times 6.94}{a^3 \times 6.023 \times 10^{23}}$ $a^3 = 4.34 \times 10^{-23}$ $\text{The edge length}$ $a = 3.52 \times 10^{-8} \text{ cm} = 352 \times 10^{-12} \text{ m} = 352 \text{ pm}$ $(\therefore 1 \text{ m} = 100 \text{ cm and } 1 \text{ pm} = 10^{-12} \text{ m})$ $\text{Hence, option 1 is the correct answer.}$

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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
}