Import Question JSON

Current Question (ID: 20831)

Question:

$50 \text{ ml of } 0.5 \text{ M oxalic acid is completely Neutralised by } 25 \text{ ml of NaOH solution.}$ $\text{Calculate the amount of NaOH (in gm) present in } 25 \text{ ml of given NaOH solution.}$ $1. \ 5$ $2. \ 2$ $3. \ 6$ $4. \ 9$

Options:

1. $5$
2. $2$
3. $6$
4. $9$

Solution:

$\text{Hint: } M_1V_1N_1 = M_2V_2N_2$ $M_1V_1N_1 = M_2V_2N_2$ $(50)(0.5)(2) = (M_2)(25)(1)$ $M_2 = 2$ $\text{Moles of NaOH } = \frac{2 \times 25}{1000} = \frac{1}{20}$ $\text{Mass of NaOH } = \frac{1}{20} \times 40 = 2 \text{ gm}$

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
}