Import Question JSON

Current Question (ID: 20824)

Question:

$\text{Molality of aqueous solution of urea is } 4.44 \text{ m. Then mole fraction of urea is } x \times 10^{-3}. \text{ The value of } x \text{ is:}$

Options:

1. $67$
2. $74$
3. $93$
4. $87$

Solution:

$\text{Hint: Mole fraction of binary solution, } X_A + X_B = 1$ $\frac{X_B}{X_A} = \frac{m \times M_A}{1000}$ $\text{Where } m = \text{molality}$ $M_A = \text{molar mass of solvent}$ $\frac{X_B}{1 - X_B} = \frac{4.44 \times 18}{1000}$ $\frac{X_B}{1 - X_B} = 0.08$ $X_B = 0.08 - 0.08 X_B$ $1.08 \times X_B = 0.08$ $X_B = 0.074 = 74 \times 10^{-3}$ $X = 74$

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
}