Import Question JSON

Current Question (ID: 8532)

Question:

$\text{At room temperature, MY and NY}_3, \text{ two nearly insoluble salts, have the same K}_{\text{sp}} \text{ values of } 6.2 \times 10^{-13}. \text{ The true statement regarding MY and NY}_3 \text{ is:}$

Options:

1. $\text{The molar solubility of MY in water is less than that of NY}_3.$
2. $\text{The salts MY and NY}_3 \text{ are more soluble in } 0.5 \text{ M KY than in pure water.}$
3. $\text{The addition of the salt of KY to a solution of MY and NY}_3 \text{ will have no effect on their solubilities.}$
4. $\text{The molar solubilities of MY and NY}_3 \text{ in water are identical.}$

Solution:

$\text{Hint: Application of Solubility product in equilibrium}\n\n\text{Step 1:}\n\n\text{Calculate the solubility of MY.}\n\n\text{MY} \rightleftharpoons \text{M}^{\text{n}+} + \text{Y}^{\text{n}-}\n\n\text{K}_{\text{sp}} = \text{s}^2\n\n\text{s} = \sqrt{\text{K}_{\text{sp}}}\n\n= \sqrt{6.2 \times 10^{-13}}\n\n\approx 7.87 \times 10^{-7}\n\n\text{Step 2:}\n\n\text{Calculate the solubility of NY}_3 \text{ is as follows:}\n\n\text{NY}_3 \rightleftharpoons \text{N}^{3+} + 3\text{Y}^{-}\n\n\text{K}_{\text{sp}} = (\text{s}) (3\text{s})^3\n\n= 27\text{s}^4\n\n\text{s} = \sqrt[4]{\frac{\text{K}_{\text{sp}}}{27}}\n\n= \sqrt[4]{\frac{6.2 \times 10^{-13}}{27}}\n\n\approx 6.88 \times 10^{-4}\n\n\text{The molar solubility of NY}_3 \text{ is more than MY. Hence, option 1 is the correct answer.}$

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
}