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Current Question (ID: 7370)

Question:

$100 \text{ mL of H}_2\text{SO}_4 \text{ solution having molarity of 1M and density 1.5 g/mL is mixed with 400 mL of water. The molarity of the H}_2\text{SO}_4 \text{ solution will be: (final density is 1.25 g/mL)}$

Options:

1. $4.4 \text{ M}$
2. $0.145 \text{ M}$
3. $0.52 \text{ M}$
4. $0.227 \text{ M}$

Solution:

$\text{Hint: Molarity} = \frac{\text{no. of moles of solute}}{\text{vol of solution in litre}}$ $\text{Step I:}$ $\text{The formula of molarity}$ $\text{Molarity} = \frac{\text{no. of moles of solute}}{\text{vol of solution in litre}}$ $\text{Step II:}$ $\text{Total moles of H}_2\text{SO}_4 = 1\text{M} \times 0.1 \text{ L} = 0.1 \text{ moles}$ $\text{The amount of H}_2\text{SO}_4 \text{ is as follows:}$ $\text{Total volume} \times \text{density} = \text{amount of H}_2\text{SO}_4$ $100 \times 1.5 = \text{amount of H}_2\text{SO}_4$ $\text{amount of H}_2\text{SO}_4 = 150 \text{ g}$ $\text{Step 3:}$ $\text{Total volume} = \frac{\text{Total mass}}{\text{Density}}$ $\text{Volume of solution} = \frac{150+400}{1.25}$ $= \frac{550}{1.25} = 440 \text{ mL}$ $\therefore \text{M} = \frac{0.1}{440} \times 1000$ $\text{M} = \frac{1}{4.4} = 0.227 \text{ M}$

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
}