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Current Question (ID: 20807)

Question:
$1.86 \text{ g of aniline completely reacts to form acetanilide.}$ $10\% \text{ of the product is lost during purification.}$ $\text{The amount of acetanilide obtained after purification (in g) is} \; \_\_\_\_ \times 10^{-2}.$
Options:
  • 1. 248
  • 2. 243
  • 3. 256
  • 4. 262
Solution:
$\text{Given } 1.86 \text{ g}$ $\Rightarrow 1 \text{ mol } \text{C}_6\text{H}_5\text{NH}_2 \text{ gives } 1 \text{ mol } \text{C}_6\text{H}_5\text{NH}\text{C}\text{CH}_3$ $\therefore \text{moles of } \text{C}_6\text{H}_5\text{NH}_2 = \text{moles of } \text{C}_6\text{H}_5\text{NH}\text{C}\text{CH}_3$ $\Rightarrow \frac{1.86}{93} = \frac{W_{\text{acetanilide}}}{135}$ $\Rightarrow W_{\text{acetanilide}} = \frac{1.86 \times 135}{93} \text{ g} = 2.70 \text{ g}$ $\text{But efficiency of reaction is } 90\%$ $\therefore \text{Mass of acetanilide produced} = 2.70 \times \frac{90}{100}$ $= 2.43 \text{ g}$ $= 243 \times 10^{-2} \text{ g}$ $\Rightarrow x = 243$

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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
}