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Current Question (ID: 7438)

Question:

$\text{Concentrated aqueous sulphuric acid is 98\% H}_2\text{SO}_4\text{ by mass and has a density}$ $\text{of 1.80 g mL}^{-1}\text{. The volume of acid required to make one litre of 0.1 M}$ $\text{H}_2\text{SO}_4\text{ solution is:}$

Options:

1. $11.10 \text{ mL}$
2. $16.65 \text{ mL}$
3. $22.20 \text{ mL}$
4. $5.55 \text{ mL}$

Solution:

$\text{Hint: Molarity} = \frac{\text{no of moles}}{\text{Volume of the solution in L}}$ $\text{Step 1: Calculate the molarity of the concentrated H}_2\text{SO}_4$ $\text{Given information:}$ $\text{- Concentrated H}_2\text{SO}_4\text{ is 98\% by mass}$ $\text{- Density = 1.80 g mL}^{-1}$ $\text{For 100 g of concentrated acid:}$ $\text{- Mass of H}_2\text{SO}_4 = 98 \text{ g}$ $\text{- Volume of solution} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.80 \text{ g/mL}} = 55.56 \text{ mL} = 0.05556 \text{ L}$ $\text{Number of moles of H}_2\text{SO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{98 \text{ g}}{98.08 \text{ g/mol}} = 0.9992 \text{ mol}$ $\text{Molarity of concentrated acid} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{0.9992 \text{ mol}}{0.05556 \text{ L}} = 18.0 \text{ M}$ $\text{Step 2: Calculate the volume needed using the dilution equation}$ $M_1V_1 = M_2V_2$ $\text{Where:}$ $M_1 = 18.0 \text{ M (concentrated acid)}$ $V_1 = \text{volume needed (to be calculated)}$ $M_2 = 0.1 \text{ M (diluted solution)}$ $V_2 = 1 \text{ L (final volume)}$ $18.0 \times V_1 = 0.1 \times 1000 \text{ mL}$ $18.0 \times V_1 = 100$ $V_1 = \frac{100}{18.0} = 5.55 \text{ mL}$ $\text{Therefore, option 4 is the correct answer.}$

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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
}