Import Question JSON

Current Question (ID: 10645)

Question:

$\text{a. CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{HF}$ $\text{b. 2HI + H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$ $\text{c. Cu + 2H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}$ $\text{d. NaCl + H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$ $\text{In which of the above-mentioned reactions does conc. H}_2\text{SO}_4\text{ act as an oxidising reagent?}$

Options:

1. $\text{a and b}$
2. $\text{b and c}$ (Correct)
3. $\text{c and d}$
4. $\text{a and d}$

Solution:

$\text{HINT: In the above given four reactions, (b) and (c) represent oxidising behaviour of H}_2\text{SO}_4\text{.}$ $\text{Explanation:}$ $\text{As we know that oxidising agent reduces itself as oxidation state of central atom decreases. Here,}$ $\text{(b) 2HI + H}_2\text{S}^{+6}\text{O}_4 \rightarrow \text{I}_2 + \text{S}^{+4}\text{O}_2 + 2\text{H}_2\text{O}$ $\text{(c) Cu}^{0} + 2\text{H}_2\text{S}^{+6}\text{O}_4 \rightarrow \text{Cu}^{+2}\text{SO}_4 + \text{S}^{+4}\text{O}_2 + 2\text{H}_2\text{O}$ $\text{In other two reactions, it is not acting as an oxidising agent as the oxidation state of all the elements remains same in reactants as well as in products.}$

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
}