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Current Question (ID: 20829)

Question:
$\text{The number of atoms in a silver plate having area } 0.05 \text{ cm}^2 \text{ and thickness } 0.05 \text{ cm is } x \times 10^{19}. \text{ The value of } x \text{ is:}$ $\text{(Density of silver is } 7.9 \text{ g/cm}^3)$
Options:
  • 1. $88$
  • 2. $90$
  • 3. $11$
  • 4. $14$
Solution:
$\text{Hint: Volume = Area } \times \text{ Thickness}$ $\text{Volume = Area } \times \text{ Thickness}$ $= 0.05 \times 0.05 \text{ cm}^3$ $= 0.0025 \text{ cm}^3$ $\text{Mass of silver } = 7.9 \times 0.0025 \text{ g}$ $\text{Moles of silver } = \frac{7.9 \times 0.0025}{108}$ $\text{Number of silver atoms } = \frac{7.9 \times 0.0025}{108} \times 6.022 \times 10^{23}$ $\Rightarrow \text{Number of silver atoms } = 0.001101 \times 10^{23}$ $= 11.01 \times 10^{19}$

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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
}