Import Question JSON

Current Question (ID: 21140)

Question:

$\text{The pH of } 0.001 \text{ M NaOH solution will be:}$

Options:

1. $3$
2. $11$
3. $-3$
4. $2$

Solution:

$\text{Hint: pH} = 14 - \text{pOH}$ $\text{Step 1:}$ $\text{The concentration of NaOH is } 0.001 \text{ M.}$ $1 \text{ mole of NaOH produces, 1 mole of OH}^- \text{ ions. Thus, the concentration of OH}^- \text{ ion is } 0.001 \text{ M.}$ $\text{Calculate the pOH of the solution as follows:}$ $\text{pOH} = -\log[\text{OH}^-]$ $= -\log[0.001]$ $\text{pOH} = 3$ $\text{Step 2:}$ $\text{pH} = 14 - \text{pOH}$ $= 14 - 3$ $= 11$

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
}