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Current Question (ID: 18070)

Question:

$100 \text{ g of liquid A (molar mass } 140 \text{ g mol}^{-1}) \text{ was dissolved in } 1000 \text{ g of liquid B (molar mass } 180 \text{ g mol}^{-1}).$ $\text{The vapour pressure of pure liquid B was found to be } 500 \text{ torr.}$ $\text{If the total vapour pressure of the solution is } 475 \text{ torr, the vapour pressure of pure liquid A will be:}$

Options:

1. $326 \text{ torr}$
2. $226 \text{ torr}$
3. $360.7 \text{ torr}$
4. $280.7 \text{ torr}$

Solution:

$\text{Hint: Use Raoult's Law}$ $\text{Explanation:}$ $\text{Step 1: Number of moles of liquid A, } n_A = \frac{100}{140} \text{ mol} = 0.714 \text{ mol}$ $\text{Number of moles of liquid B, } n_B = \frac{1000}{180} \text{ mol} = 5.556 \text{ mol}$ $\text{Then, the mole fraction of A, } x_A = \frac{n_A}{n_A + n_B}$ $= \frac{0.714}{0.714 + 5.556}$ $= 0.114$ $\text{And, mole fraction of B, } x_B = 1 - 0.114 = 0.886$ $\text{Step 2: Vapour pressure of pure liquid B, } P_B^0 = 500 \text{ torr}$ $\text{Therefore, vapour pressure of liquid B in the solution,}$ $P_B = P_B^0 x_B$ $= 500 \times 0.886$ $= 443 \text{ torr}$ $\text{Total vapour pressure of the solution, } P_{\text{total}} = 475 \text{ torr}$ $\therefore \text{Vapour pressure of liquid A in the solution,}$ $P_A = P_{\text{total}} - P_B = 475 - 443 = 32 \text{ torr}$ $\text{Step 3: Now,}$ $P_A = P_A^0 x_A$ $\Rightarrow P_A^0 = \frac{P_A}{x_A} = \frac{32}{0.114}$ $= 280.7 \text{ torr}$ $\text{Hence, the vapour pressure of pure liquid A is } 280.7 \text{ torr.}$

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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
}