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Current Question (ID: 10663)

Question:

$\text{Which of the following statements is true?}$

Options:

1. $\text{S} - \text{S bond is present in } \text{H}_2\text{S}_2\text{O}_6$ (Correct)
2. $\text{In peroxosulphuric acid } (\text{H}_2\text{SO}_5) \text{ sulphur is in } +5 \text{ oxidation state.}$
3. $\text{Iron powder, along with } \text{Al}_2\text{O}_3 \text{ and } \text{K}_2\text{O}, \text{ is used as a catalyst in the preparation of } \text{NH}_3 \text{ by Haber's process.}$
4. $\text{Change in enthalpy is positive for the preparation of } \text{SO}_3 \text{ by catalytic oxidation of } \text{SO}_2$

Solution:

$\text{HINT- In } \text{H}_2\text{SO}_5, \text{ oxidation state of S is } +6$ $\text{Explanation:}$ $\text{STEP1-}$ $\text{(a) Structure of } \text{H}_2\text{S}_2\text{O}_6 \text{ is as shown below:}$ $\text{It contains one S-S bond.}$ $\text{(b) In peroxosulphuric acid } (\text{H}_2\text{SO}_5) \text{ sulphur is in } +6 \text{ oxidation state.}$ $\text{Structure of } \text{H}_2\text{SO}_5 \text{ is :}$ $\text{Let oxidation state of S = x:}$ $2 \times (+1) + x + 3 \times (-2) + 2 \times (-1) = 0$ $x - 6 = 0$ $x = 6$ $\text{STEP2- (c) During preparation of ammonia, iron oxide with small amount of } \text{K}_2\text{O} \text{ and } \text{Al}_2\text{O}_3 \text{ is used as a catalyst to increase the rate of attainment of equilibrium.}$ $\text{(d) Change in enthalpy is negative for preparation of } \text{SO}_3 \text{ by catalytic oxidation of } \text{SO}_2.$

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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
}