Import Question JSON

Current Question (ID: 21431)

Question:

$\text{A sodium chloride (NaCl) solution has a molarity of 3 M. Given that the density of the solution is 1.25 g/mL, its molality will be:}$

Options:

1. $2.90$
2. $2.79$
3. $1.85$
4. $3.85$

Solution:

$\text{Given molarity of solution} = 3 \text{ M means 3 moles of NaCl is present in 1000 mL of solution.}$ $\text{Mass of solution} = d \times v$ $= 1.25 \times 1000$ $= 1250 \text{ g}$ $\text{Mass of solute} = 3 \times 58.5 = 175.5 \text{ g}$ $\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}$ $= 1250 - 175.5 = 1074.5 \text{ g} = 1.0745 \text{ kg}$ $\text{Molality}$ $m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$ $m = \frac{3}{1.0745} = 2.79 \text{ m}$

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
}