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Current Question (ID: 17946)

Question:

$\text{Lithium borohydride (LiBH}_4\text{), crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are:}$ $a = 6.81 \, \text{\AA}, \ b = 4.43 \, \text{\AA}, \ c = 7.17 \, \text{\AA}. \text{If the molar mass of LiBH}_4 \text{ is } 21.76 \, \text{g mol}^{-1}, \text{ then the density of the crystal is:}$

Options:

1. $0.67 \, \text{g cm}^{-3}$
2. $0.58 \, \text{g cm}^{-3}$
3. $1.23 \, \text{g cm}^{-3}$
4. $\text{None of the above.}$

Solution:

$\text{Density;} \, \rho = \frac{ZM}{NV}$ $\text{Calculate the density of crystal is as follows:}$ $\rho = \frac{ZM}{NV}$ $= \frac{4 \times (21.76) \, \text{g mol}^{-1} (6.023 \times 10^{23})}{\text{mol}^{-1} (6.81 \times 4.43 \times 7.17 \times 10^{-24}) \, \text{cm}^3}$ $= 0.668 \, \text{g cm}^{-3}$

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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
}