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Current Question (ID: 7423)

Question:

$\text{The molality of pure water is:}$

Options:

1. $1 \text{ m}$
2. $18 \text{ m}$
3. $55.5 \text{ m}$
4. $\text{None of the above}$

Solution:

$\text{Hint: Molarity = number of moles of solute / Volume of solution in L}$ $\text{Explanation:}$ $\text{The density of pure water at room temperature i.e., } 25 ^{\circ}\text{C is } 0.9970749 \text{ g/ml, therefore the mass will be } 0.9970749 \text{ kg, and the molecular mass is } 18.0148 \text{ g/mol. The molality is } 55.510 \text{ m.}$ $\text{At that amount of water, the number of moles is:}$ $\frac{997.0749 \text{ g}}{18.0148 \text{ g/mol}} = 55.348 \text{ mole}$ $\text{Molality}$ $\frac{\text{mol water}}{\text{kg water}} = \frac{55.348 \text{ mol}}{0.997 \text{ kg}} = 55.10 \text{ m}$

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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
}