Import Question JSON

Current Question (ID: 20835)

Question:

$\text{If 84g of NaOH (aq) is present in a 3 molar solution, find volume of solution (in mL).}$ $\text{1. 800}$ $\text{2. 700}$ $\text{3. 400}$ $\text{4. 500}$

Options:

1. $800$
2. $700$
3. $400$
4. $500$

Solution:

$\text{Hint: Molarity} = \frac{\text{Moles of NaOH}}{\text{Volume of solution in L}}$ $\text{As we know}$ $\text{Molarity} = \frac{\text{Moles of NaOH}}{\text{Volume of solution in L}} = \frac{84}{\text{M.W. of NaOH} \times \text{Volume of Solution (L)}}$ $3\text{M} = \frac{84}{40 \times 3}$ $\text{Volume of Solution} = \frac{84}{40 \times 3} \text{ L} = 0.7 \text{ L}$ $\text{Volume of Solution in mL} = 0.7 \times 1000 = 700 \text{ mL}$

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
}