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Current Question (ID: 9154)

Question:
$\text{The pair of compounds, that cannot exist together in a solution is -}$
Options:
  • 1. $\text{NaHCO}_3 \text{ and NaOH}$
  • 2. $\text{Na}_2\text{CO}_3 \text{ and NaOH}$
  • 3. $\text{NaHCO}_3 \text{ and Na}_2\text{CO}_3$
  • 4. $\text{NaHCO}_3 \text{ and H}_2\text{O}$
Solution:
$\text{HINT: Compounds that react with each other can't exist together in the solution.}$ $\text{Explanation:}$ $\text{NaHCO}_3 \text{ and NaOH react to form Na}_2\text{CO}_3 \text{ \& H}_2\text{O because NaHCO}_3 \text{ is an acidic salt \& NaOH release. Other pairs do not react to each other.}$ $\text{NaHCO}_3 + \text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O}$ $\text{So, NaHCO}_3 \text{ \& NaOH cannot exist together.}$

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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
}