Import Question JSON

Current Question (ID: 21432)

Question:

$\text{If the density of an } x \text{ M solution of NaOH is } 1.12 \text{ g/mL and its}$ $\text{molality is } 3 \text{ m, determine the value of } x.$

Options:

1. $3.0$
2. $2.3$
3. $4.0$
4. $5.2$

Solution:

$\text{Given molality of NaOH} = 3 \text{ m}$ $\text{It means 3 moles of NaOH present in 1000 g of solvent.}$ $\text{Mass of solute (NaOH)} = 3 \times 40 = 120 \text{ g}$ $\text{Mass of solution} = 1000 + 120 = 1120 \text{ g}$ $\text{Density of solution} = 1.12 = \frac{1120}{\text{volume}}$ $\text{Volume of solution} = \frac{1120}{1.12} = 1000 \text{ mL}$ $\text{Molarity of solution} = \frac{3}{1000} \times 1000$ $= 3 \text{M}$

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
}