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Current Question (ID: 8139)

Question:
$\text{The density of }\text{O}_2\text{(g) is maximum at-}$
Options:
  • 1. $\text{STP}$
  • 2. $273\ \text{K and } 2\ \text{atm}$
  • 3. $546\ \text{K and } 1\ \text{atm}$
  • 4. $546\ \text{K and } 2\ \text{atm}$
Solution:
$\text{Hint: Density} = \frac{\text{PM}}{\text{RT}} \text{Step 1:} \text{The formula of density is as follows:} \text{Density} = \frac{\text{PM}}{\text{RT}} \text{Step 2:} \text{From the formula of density it is clear that d is directly proportional to pressure and inversely proportional to temperature.} $\text{d} = \text{PM/RT}$ $\text{d} \propto \text{P/T for } \text{d}_{\text{max}}, \text{ P/T should be max}\uparrow$ \text{Hence, when temperature is 273 K and pressure is 2 atm. The value of density is maximum.}$

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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
}