Import Question JSON

Current Question (ID: 21016)

Question:

$\text{The internal energy change (in kJ) when 90 g of water undergoes complete evaporation at } 100^\circ\text{C is -}$ $\text{(Given: } \Delta H_{\text{vap}} \text{ for water at } 373 \text{ K} = 41 \text{ kJ/mol, } R = 8.314 \text{ J K}^{-1}\text{mol}^{-1})$

Options:

1. $154.5$
2. $168.5$
3. $176.5$
4. $189.5$

Solution:

$\text{Hint: Use formula } \Delta H = \Delta U + \Delta n_g RT$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ $\text{The formula used to calculate internal energy is as follows:}$ $\Delta H = \Delta U + \Delta n_g RT$ $\text{The given values are as follows:}$ $\Delta H_{\text{vap}} \text{ for water at } 373 \text{ K} = 41 \text{ kJ/mol, } R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}, \text{ and } \Delta n_g = 1$ $\text{Step 2:}$ $\text{Calculate the value of } \Delta U \text{ for 1 mole or 18 g of water in kJ as follows:}$ $41 \text{ kJ/mol} = \Delta U + 1 \times 8.314 \times 373$ $41 - 8.314 \times 373 = \Delta U$ $41 - 3.1 = \Delta U$ $37.9 \text{ kJ/mol} = \Delta U$ $\text{1 mole } \text{H}_2\text{O} \text{ is 18 g calculate the } \Delta U \text{ value for 90 g or 5 mol of } \text{H}_2\text{O}$ $\Delta U = 37.9 \text{ kJ/mol} \times 5$ $= 189.5 \text{ kJ}$

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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
}