Import Question JSON

Current Question (ID: 8682)

Question:

$\text{The incorrect statement among the following options is:}$

Options:

1. $\text{K}^+ \text{ ions react with alkaline water and give K(OH)}_3$
2. $\text{KCl is a salt of strong acid and strong base.}$
3. $\text{Al(OH)}_3 \text{ in alkaline condition gives [Al(OH)}_4\text{]}^-$
4. $\text{AlCl}_3 \text{ is a salt of strong acid and weak base.}$

Solution:

$\text{Hint: KCl remains as ions in acidic and alkaline conditions.}$ $\text{Explanation:}$ $\text{Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water.}$ $\text{It dissociates into ions as follows:}$ $\text{KCl}_{(s)} \rightarrow \text{K}^+\text{}_{(aq)} + \text{Cl}^-\text{}_{(aq)}$ $\text{In acidified and alkaline water, the ions do not react and remain as such.}$ $\text{Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base Al(OH)}_3$ $\text{Hence, it undergoes hydrolysis in normal water.}$ $\text{AlCl}_3\text{}_{(s)} + 3\text{H}_2\text{O}_{(l)} \rightarrow \text{Al(OH)}_3\text{}_{(s)} + 3\text{H}^+\text{}_{(aq)} + 3\text{Cl}^-\text{}_{(aq)}$ $\text{In acidified water, H}^+ \text{ ions react with Al(OH)}_3 \text{ forming water and giving Al}^{3+} \text{ ions. Hence, in acidified water, will exist as Al}^{3+} \text{ and Cl}^- \text{ ions.}$ $\text{Al(OH)}_3\text{}_{(s)} \rightarrow \text{Al}^{3+}\text{}_{(aq)} + 3\text{Cl}^-\text{}_{(aq)}$ $\text{In alkaline water the following reaction takes place:}$ $\text{Al(OH)}_3\text{}_{(s)} + \text{OH}^-\text{}_{(aq)} \rightarrow [\text{Al(OH)}_4]^-\text{}_{(aq)} + 2\text{H}_2\text{O}_{(l)} \text{ From alkaline water}$

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
}