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Current Question (ID: 21402)

Question:

$\text{At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. The vapour pressure in mm Hg of n-heptane in its pure state is:}$ $1. \ 500$ $2. \ 600$ $3. \ 700$ $4. \ 800$

Options:

1. $500$
2. $600$
3. $700$
4. $800$

Solution:

$\text{Hint: The formula of total pressure is } P_T = P_A + P_B$ $\text{Step 1:}$ $\text{Let A is n-hexane and B is n-heptane.}$ $\text{The total pressure is } P_T = P_A + P_B$ $\text{The total pressure is 550 mm of Hg when the solution contains 1 mole of n-hexane and 3 moles of n-heptane.}$ $\text{The formula of partial pressure is as follows:}$ $P_A = P_A^0 X_A$ $P_B = P_B^0 X_B$ $550 = P_A^0 \times \frac{1}{4} + P_B^0 \times \frac{3}{4}$ $2200 = P_A^0 + 3P_B^0 \quad \text{......(1)}$ $\text{Step 2:}$ $\text{When one more mole of n-heptane is added to this solution, the vapor pressure of the solution increases by 10 mm of Hg.}$ $560 = P_A^0 \times \frac{1}{5} + P_B^0 \times \frac{4}{5}$ $2800 = P_A^0 + 4P_B^0 \quad \text{......(2)}$ $\text{Step 3:}$ $\text{Equate equations 1 and 2 as follows:}$ $2200 = P_A^0 + 3P_B^0$ $2800 = P_A^0 + 4P_B^0$ $P_B^0 = 600$ $2800 = P_A^0 + 4(600)$ $2800 - 2400 = P_A^0$ $P_A^0 = 400$ $P_A^0 = 400$ $P_B^0 = 600$

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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
}