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Current Question (ID: 20798)

Question:

$\text{If } 0.02 \text{ moles of } \text{H}_2\text{SO}_4 \text{ are present in 1 litre of solution from which 50\% solution is taken out and again diluted up to 1 litre by adding water. Further, if } 0.01 \text{ moles of } \text{H}_2\text{SO}_4 \text{ are added, then the total millimoles of } \text{H}_2\text{SO}_4 \text{ in the resulting solution will be:}$ $1.\ 10$ $2.\ 5$ $3.\ 20$ $4.\ 15$

Options:

1. $10$
2. $5$
3. $20$
4. $15$

Solution:

$\text{Hint: On dilution, the number of moles of substance does not change}$ $\text{Step 1:}$ $\text{Initial moles of } \text{H}_2\text{SO}_4 \text{ (in/Lit.) solution } = 0.02 \text{ mol}$ $\text{When 50\% } \text{H}_2\text{SO}_4 \text{ solutions are taken out then a mole of } \text{H}_2\text{SO}_4 \text{ remains in the solution is } 0.01 \text{ mol.}$ $\text{Then 1 litre of water and } 0.01 \text{ mol of } \text{H}_2\text{SO}_4 \text{ again added. The final mole of } \text{H}_2\text{SO}_4 \text{ in the solution is } 0.02 \text{ mol.}$ $\text{Step 2:}$ $\text{Calculate the moles in millimole as follows:}$ $\text{The final mole of } \text{H}_2\text{SO}_4 \text{ in the solution is } 0.02 \text{ mol.}$ $1 \text{ mole } = 1000 \text{ millimole}$ $= 0.02 \text{ mol } \times \frac{1000 \text{ millimole}}{1 \text{ mol}}$ $= 20 \text{ millimoles}$

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
}