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Current Question (ID: 8183)

Question:

$\text{A gaseous mixture contains 56 g of } \text{N}_2\text{, 44 g of } \text{CO}_2 \text{ and 16 g of } \text{CH}_4\text{. The total pressure of the mixture is 720 mm Hg. The partial pressure of } \text{CH}_4 \text{ is-}$

Options:

1. $180 \text{ mm}$
2. $360 \text{ mm}$
3. $540 \text{ mm}$
4. $720 \text{ mm}$

Solution:

$\text{Hint: Dalton Law of partial pressure}$\n\n$\text{Step 1:}$\n\n$\text{The formula of Dalton Law of partial pressure is as follows:}$\n\n$\text{P(partial pressure)} = \text{X}_{\text{gas}}\text{P}_\text{T}$\n\n$\text{Calculate the moles of } \text{N}_2\text{, } \text{CO}_2\text{, and } \text{CH}_4 \text{ as follows:}$\n\n$\text{The formula of number of mole is as follows:}$\n\n$\text{number of mole} = \frac{\text{amount of substance}}{\text{molar mass of substance}}$\n\n$\text{n}_{\text{N}_2} = \frac{56}{28}$\n\n$= 2 \text{ mol}$\n\n$\text{n}_{\text{CO}_2} = \frac{44}{44}$\n\n$= 1 \text{ mol}$\n\n$\text{n}_{\text{CH}_4} = \frac{16}{16}$\n\n$= 1 \text{ mol}$\n\n$\text{Step 2:}$\n\n$\text{Calculate the partial pressure of } \text{CH}_4 \text{ is as follows:}$\n\n$\text{P(partial pressure)} = \text{X}_{\text{CH}_4}\text{P}_\text{T}$\n\n$\text{P}(\text{CH}_4) = \frac{16/16}{56/28+44/44+16/16} \times 720$\n\n$= \frac{1}{2+1+1} \times 720$\n\n$= \frac{1}{4} \times 720$\n\n$= 180 \text{ mm}$

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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
}