Import Question JSON

Current Question (ID: 21065)

Question:

$\text{Given below are two statements:}$ $\text{Assertion (A):}$ $\text{When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid.}$ $\text{Reason (R):}$ $\text{The equilibrium constant of Cu}^{2+}\text{(aq)} + \text{S}^{2-}\text{(aq)} \rightleftharpoons \text{CuS(s)} \text{ is high because the solubility product is low.}$

Options:

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

Solution:

$\text{Both assertion and reason are correct but the reason is not a correct explanation of assertion.}$ $\text{Because the rate of the reaction tells about the kinetics of the reaction but it does not tell about the extent of reaction.}$ $\text{Hence, the rate of chemical reaction has nothing to do with the value of the equilibrium constant.}$ $\text{In the reason, it is given that the reaction has high equilibrium constant value.}$ $\text{It indicate that CuS is highly stable compound.}$ $\text{The solubility product of CuS is low it indicate that when aqueous solution of CuS is prepared less amount of CuS will dissociate and form Cu}^{2+}\text{ and S}^{2-}\text{ ions.}$

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
}