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Current Question (ID: 21027)

Question:

$\text{Enthalpy of sublimation of iodine is } 24 \text{ cal g}^{-1} \text{ at } 200 \, ^\circ\text{C. If specific heat of } \text{I}_2(\text{s}) \text{ and } \text{I}_2(\text{vap}) \text{ are } 0.055 \text{ and } 0.031 \text{ cal g}^{-1}\text{K}^{-1} \text{ respectively, then enthalpy of sublimation of iodine at } 250^\circ\text{C in cal g}^{-1} \text{ is:}$

Options:

1. $2.85$
2. $22.8$
3. $11.4$
4. $5.7$

Solution:

$\text{Hint: Use the relation between } \Delta H = C_p \Delta T$ $\text{The relation between } \Delta H \text{ and } C_p \text{ is as follows:}$ $\Delta H = C_p \Delta T$ $\text{For two different temperature conditions the formula is as follows:}$ $\Delta H_{T_2} - \Delta H_{T_1} = \Delta C_p (T_2 - T_1)$ $\text{I}_2(\text{s}) \rightarrow \text{I}_2(\text{g}): \Delta H_1 = 24 \text{ cal g}^{-1} \text{ at } 200 \, ^\circ\text{C}$ $\Delta H_2 = \Delta H_1 + \Delta C_p (T_2 - T_1)$ $= 24 + (0.031 - 0.055) \times 50$ $= 24 - 1.2$ $= 22.8 \text{ cal g}^{-1}$

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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
}