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Current Question (ID: 8534)

Question:

\text{The minimum volume of water required to dissolve } 1\text{g of calcium sulphate at } 298 \text{ K is} (\text{For } \text{CaSO}_4, \text{ K}_{\text{sp}} \text{ is } 9.1 \times 10^{-6})

Options:

1. $1.22 \text{ L}$
2. $0.69 \text{ L}$
3. $2.44 \text{ L}$
4. $1.87 \text{ L}$

Solution:

\text{Hint: K}_{\text{sp}} = \text{s}^2 \text{Explanation:} \text{Step 1:} \text{CaSO}_{4(s)} \rightleftharpoons \text{Ca}^{2+}_{(aq)} + \text{SO}_4^{2-}_{(aq)} \text{K}_{\text{sp}} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \text{K}_{\text{sp}} = \text{s}^2 \text{, where "s" is the solubility of CaSO}_4 9.1 \times 10^{-6} = \text{s}^2 \Rightarrow \text{s} = 3.02 \times 10^{-3} \text{ mol L}^{-1} \text{Step 2:} \text{Molecular mass of CaSO}_4 = 136 \text{ g mol}^{-1} \text{Solubility of CaSO}_4 \text{ in g L}^{-1} = 3.02 \times 10^{-3} \times 136 = 0.41 \text{ g L}^{-1} \text{So, 1 L water is required to dissolve 0.41 g of CaSO}_4 \text{Thus to dissolve 1 g of CaSO}_4 \text{ we require } = \frac{1}{0.41} = 2.44 \text{ L of water}

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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
}