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Current Question (ID: 17926)

Question:

$\text{Silver crystallizes in fcc lattice. If the edge length of the cell is } 4.07 \times 10^{-8} \text{ cm}$ $\text{and density is } 10.5 \text{ g cm}^{-3}, \text{ the atomic mass of silver is -}$

Options:

1. $120.31 \text{ g mol}^{-1}$
2. $107.9 \text{ g mol}^{-1}$
3. $207.60 \text{ g mol}^{-1}$
4. $138.52 \text{ g mol}^{-1}$

Solution:

$\text{HINT: } \rho = \frac{Z \times M}{N_A \times a^3}$ $\text{STEP 1:}$ $\text{It is given that the edge length, } a = 4.07 \times 10^{-8} \text{ cm}$ $\text{Density, } d = 10.5 \text{ g cm}^{-3}$ $\text{As the lattice is fcc type, the number of atoms per unit cell, } z = 4$ $\text{STEP 2:}$ $\rho = \frac{z \times M}{N_A \times a^3}$ $\Rightarrow M = \frac{\rho \times N_A \times a^3}{z}$ $\Rightarrow M = \frac{10.5 \times (4.07 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4}$ $\Rightarrow M = 107.9 \text{ g mol}^{-1}$ $\text{The atomic mass of silver is } 107.9 \text{ g mol}^{-1}$

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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
}