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Current Question (ID: 18041)

Question:

$\text{The partial pressure of ethane over a solution containing } 6.56 \times 10^{-2} \text{ g of ethane is } 1 \text{ bar.}$ $\text{If the solution contains } 5.00 \times 10^{-2} \text{ g of ethane, the partial pressure of the gas will be:}$

Options:

1. $0.66 \text{ bar}$
2. $0.96 \text{ bar}$
3. $0.76 \text{ bar}$
4. $0.19 \text{ bar}$

Solution:

$\text{Hint: Use Henry's law}$ $\text{Explanation:}$ $\text{Step 1: Molar mass of ethane (C}_2\text{H}_6) = 2 \times 12 + 6 \times 1 = 30 \text{ g mol}^{-1}$ $\therefore \text{Number of moles present in } 6.56 \times 10^{-2} \text{ g of ethane} = \frac{6.56 \times 10^{-2}}{30} = 2.187 \times 10^{-3} \text{ mol}$ $\text{Step 2: According to Henry's law,}$ $p = K_H x$ $\Rightarrow 1 \text{ bar} = K_H \cdot \frac{2.187 \times 10^{-3}}{2.187 \times 10^{-3} + x}$ $\Rightarrow 1 \text{ bar} = K_H \cdot \frac{x}{2.187 \times 10^{-3}}$ $\Rightarrow K_H = \frac{x}{2.187 \times 10^{-3}} \text{ bar (since } x > 2.187 \times 10^{-3})$ $\text{Step 3: Number of moles present in } 5.00 \times 10^{-2} \text{ g of ethane} = \frac{5.00 \times 10^{-2}}{30} = 1.67 \times 10^{-3} \text{ mol}$ $\text{According to Henry's law.}$ $p = K_H x$ $= \frac{x}{2.187 \times 10^{-3}} \times \frac{1.67 \times 10^{-3}}{(1.67 \times 10^{-3}) + x}$ $= \frac{2.187 \times 10^{-3}}{2.187 \times 10^{-3}} \times \frac{1.67 \times 10^{-3}}{x}$ $= 0.764$ $\text{Hence, the partial pressure of the gas shall be } 0.764 \text{ bar.}$

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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
}