Import Question JSON

Current Question (ID: 18864)

Question:

$\text{Match the solutions given in Column I with the colours given in Column II.}$ $\text{Column I (Aqueous solution of salt)}$ $\begin{array}{ll} \text{A. FeSO}_4 \cdot \text{7H}_2\text{O} & \text{1. Green} \\ \text{B. NiCl}_2 \cdot \text{4H}_2\text{O} & \text{2. Light pink} \\ \text{C. MnCl}_2 \cdot \text{4H}_2\text{O} & \text{3. Blue} \\ \text{D. CoCl}_2 \cdot \text{6H}_2\text{O} & \text{4. Pale green} \\ & \text{5. Pink} \end{array}$ $\text{Codes:}$ $\begin{array}{cccc} \text{A} & \text{B} & \text{C} & \text{D} \\ 2 & 3 & 4 & 1 \\ 1 & 2 & 3 & 5 \\ 5 & 4 & 3 & 2 \\ 4 & 1 & 2 & 5 \end{array}$

Options:

1. $2, 3, 4, 1$
2. $1, 2, 3, 5$
3. $5, 4, 3, 2$
4. $4, 1, 2, 5$

Solution:

$\text{HINT: MnCl}_2 \cdot \text{4H}_2\text{O is light pink colour.}$ $\text{Explanation:}$ $\text{The d block complex shows colour because of the d-d transition.}$ $\begin{array}{ll} \text{An aqueous solution of salt} & \text{Colour} \\ \text{A. FeSO}_4 \cdot \text{7H}_2\text{O} & \text{Pale green} \\ \text{B. NiCl}_2 \cdot \text{4H}_2\text{O} & \text{Green} \\ \text{C. MnCl}_2 \cdot \text{4H}_2\text{O} & \text{Light pink} \\ \text{D. CoCl}_2 \cdot \text{6H}_2\text{O} & \text{Pink} \end{array}$ $\text{Hence, option fourth is the correct answer.}$

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
}