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Current Question (ID: 7430)

Question:
The total molarity of all the ions present in 0.1 M of $\text{CuSO}_4$ and 0.1 M of $\text{Al}_2(\text{SO}_4)_3$ solution is:
Options:
  • 1. 0.2 M
  • 2. 0.7 M
  • 3. 0.8 M
  • 4. 1.2 M
Solution:
\text{Hint: Molarity} = \frac{\text{number of moles}}{\text{volume of solution(L)}}\n\n\text{Explanation:}\n\n\text{Moles of } \text{Cu}^{2+} = 0.1 \times 1 = 0.1\n\n\text{Moles of } \text{SO}_4^{2-} = 0.1 \times 1 = 0.1\n\n\text{Moles of } \text{Al}^{3+} = 0.1 \times 2 = 0.2\n\n\text{Moles of } \text{SO}_4^{2-} = 0.1 \times 3 = 0.3\n\n\text{Total moles of ions present in 1 litre} = 0.7\n\n\text{Molarity of all ions} = 0.7 \text{ M}\n\n\text{Thus, option 2 is the correct answer.}

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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
}