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Current Question (ID: 17920)

Question:

$\text{CsBr crystallises in a body-centered cubic lattice. The unit cell length is } 436.6 \text{ pm.}$ $\text{Given that the atomic mass of Cs = 133 amu and that of Br = 80 amu}$ $\text{and the Avogadro number being } 6.02 \times 10^{23} \text{ mol}^{-1}, \text{ the density of CsBr is:}$

Options:

1. $42.5 \text{ g/cm}^3$
2. $0.425 \text{ g/cm}^3$
3. $8.25 \text{ g/cm}^3$
4. $4.25 \text{ g/cm}^3$

Solution:

$\text{Density} = \frac{Z \times M}{a^3 \times N_A}$ $\text{Density of any crystal type can be found out using the formula given below.}$ $\text{Density of CsBr} = \frac{Z \times M}{a^3 \times N_A}$ $Z = \text{no. of atoms in the bcc unit cell (here } Z=1 \text{ because CsBr is pseudo bcc crystal)}$ $M = \text{molar mass of CsBr (133 + 80 = 213)}$ $a = \text{edge length of unit cell (436.6 pm or } 436.6 \times 10^{-10} \text{ cm)}$ $\text{By substitution, in 1.}$ $\text{Density} = \frac{1 \times 213}{6.023 \times 10^{23} \times (436.6 \times 10^{-10})^3}$ $\text{Density} = 4.25 \text{ g/cm}^3$

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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
}