Import Question JSON

Current Question (ID: 10688)

Question:

$\text{Select the correct option based on statements below:}$ $\text{Assertion (A): HI cannot be prepared by the reaction of KI with concentrated H}_2\text{SO}_4\text{.}$ $\text{Reason (R): HI has the lowest H-X bond strength among halogen acids.}$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$ (Correct)
3. $\text{(A) is True but (R) is False.}$
4. $\text{(A) is False but (R) is True.}$

Solution:

$\text{HINT: H}_2\text{SO}_4 \text{ is a strong oxidising agent}$ $\text{Explanation:}$ $\text{STEP 1: HI cannot be prepared by the reaction of KI with concentrated H}_2\text{SO}_4 \text{ because H}_2\text{SO}_4 \text{ is a strong oxidising agent and it oxidised I}^- \text{ into I}_2\text{.}$ $\text{The reaction will be:}$ $2\text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{I}_2 + \text{H}_2\text{S} + \text{H}_2\text{O}$ $\text{STEP 2: In the case of hydrogen halides (HX), The stability of these halides decreases down the group due to a decrease in bond (H-X) dissociation enthalpy. Thus, HI has the lowest H-X bond strength among halogen acids.}$ $\text{Here, both Assertion and Reason are correct statements, and Reason is not the correct explanation of the Assertion.}$

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
}