Import Question JSON

Current Question (ID: 8537)

Question:

$\text{The solubility of } \text{BaSO}_4 \text{ in water is } 2.42 \times 10^{-3} \text{ g/ litre at } 298 \text{ K. The value of the solubility product will be: (Molar mass of } \text{BaSO}_4 = 233 \text{ g mol}^{-1})$

Options:

1. $1.08 \times 10^{-10} \text{ mol}^2 \text{ L}^{-2}$
2. $1.08 \times 10^{-12} \text{ mol}^2 \text{ L}^{-2}$
3. $1.08 \times 10^{-14} \text{ mol}^2 \text{ L}^{-2}$
4. $1.08 \times 10^{-8} \text{ mol}^2 \text{ L}^{-2}$

Solution:

$\text{Hint: } \text{K}_{sp} = \text{S}^2 \n\n\text{Step 1:} \n\n\text{The solubility value is } 2.42 \times 10^{-3} \text{ g L}^{-1}\text{, but the correct unit of solubility is mol L}^{-1}. \text{ Divide the solubility by molar mass.} \n\n\text{S} = \frac{2.42 \times 10^{-3}}{233} \n\n= 1.0386 \times 10^{-5} \approx 1.04 \times 10^{-5} \text{ mol L}^{-1} \n\n\text{Step 2:} \n\n\text{The reaction is as follows:} \n\n\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-} \n\n\text{K}_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \n\n\text{Since the stoichiometry is 1:1, if S is the molar solubility, then } [\text{Ba}^{2+}] = \text{S} \text{ and } [\text{SO}_4^{2-}] = \text{S}. \n\n\text{K}_{sp} = \text{S}^2 \n\n\text{K}_{sp} = (1.0386 \times 10^{-5})^2 \n\n= 1.07869 \times 10^{-10} \n\n\approx 1.08 \times 10^{-10}$

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
}