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Current Question (ID: 8187)

Question:

$\text{The partial pressure of } \text{H}_2 \text{ in a flask containing 2 g of } \text{H}_2\text{, 14 g of } \text{N}_2\text{, and 16 g of } \text{O}_2 \text{ is-}$

Options:

1. $1/2 \text{ of the total pressure}$
2. $1/3 \text{ of the total pressure}$
3. $1/4 \text{ of the total pressure}$
4. $1/16 \text{ of the total pressure}$

Solution:

$\text{Hint: Use Dalton's law}$\n\n$\text{Step 1:}$\n\n$\text{The formula of Dalton's law is as follows:}$\n\n$\text{P(partial pressure)} = \text{P}_\text{T}\text{X}_{\text{gas}}$\n\n$\text{Calculate the mole of all the gases as follows:}$\n\n$\text{n}(\text{H}_2) = \frac{2}{2} = 1$\n\n$\text{n}(\text{N}_2) = \frac{14}{28} = 0.5$\n\n$\text{n}(\text{O}_2) = \frac{16}{32} = 0.5$\n\n$\text{Step 2:}$\n\n$\text{Calculate the partial pressure of } \text{H}_2 \text{ as follows:}$\n\n$\text{P}(\text{H}_2) = \frac{1}{1+0.5+0.5}\text{P}_\text{T}$\n\n$\text{P}(\text{H}_2) = \frac{1}{2}\text{P}_\text{T}$

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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
}