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Current Question (ID: 7434)

Question:

The molality of a 15\% (w/vol.) solution of $\text{H}_2\text{SO}_4$ of density 1.1g/cm$^3$ is-

Options:

1. 1.2
2. 1.4
3. 1.8
4. 1.6

Solution:

\text{Hint: molality} = \frac{\text{amount of } \text{H}_2\text{SO}_4}{\text{molar mass of } \text{H}_2\text{SO}_4 \times \text{amount of solvent (kg)}}\n\n15 \text{ g } \text{H}_2\text{SO}_4 \text{ in 100 ml of solution}\n\n\text{Weight of solution} = 1.1 \times 100 = 110 \text{ gram}\n\n\text{Solvent} = 110 - 15 = 95 \text{ g}\n\n\text{molality} = \frac{\text{amount of } \text{H}_2\text{SO}_4}{\text{molar mass of } \text{H}_2\text{SO}_4 \times \text{amount of solvent (kg)}}\n\n= \frac{15}{98} \times \frac{1000}{95}\n\n= 1.6

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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
}