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Current Question (ID: 7662)

Question:

$\text{IP}_1$ and $\text{IP}_2$ for $\text{Mg}$ are $178 \text{ Kcal mol}^{-1}$ and $348 \text{ Kcal mol}^{-1}$, respectively. The energy required for the reaction, $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^{-}$ will be:

Options:

1. $+170 \text{ Kcal mol}^{-1}$
2. $+526 \text{ Kcal mol}^{-1}$ (Correct)
3. $-170 \text{ Kcal mol}^{-1}$
4. $-526 \text{ Kcal mol}^{-1}$

Solution:

HINT: Ionization potential is additive property. Explanation: The energy required for the reaction, $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^{-}$ is calculated by adding $\text{IP}_1$ and $\text{IP}_2$ The energy required $= \text{IP}_1 + \text{IP}_2$ $= 178 + 348$ $= 526 \text{ Kcal mol}^{-1}$

Import JSON File

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
}