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Current Question (ID: 21404)

Question:

$\text{The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively, at 298 K.}$ $\text{On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture.}$ $\text{The mole fraction of liquid B is 0.5 in the mixture.}$ $\text{The vapour pressure of the final solution, the mole fractions of components A and B in the vapour phase, respectively, are:}$

Options:

1. $500 \text{ mm Hg}, 0.5, 0.5$
2. $450 \text{ mm Hg}, 0.5, 0.5$
3. $500 \text{ mm Hg}, 0.4, 0.6$
4. $450 \text{ mm Hg}, 0.4, 0.6$

Solution:

$\text{Hint: Use the Raoult's law equation}$ $\text{Step 1:}$ $\text{Calculate the total pressure of the solution as follows:}$ $P_A^0 = 400 \text{ mmHg}$ $P_B^0 = 600 \text{ mmHg}$ $x_B = 0.5$ $x_A = 1 - 0.5 = 0.5$ $P_T = P_A^0 x_A + P_B^0 x_B$ $= 400(0.5) + 600(0.5)$ $= 200 + 300 = 500$ $P_T = 500 \text{ mmHg}$ $\text{Step 2:}$ $\text{Calculate the mole fraction of A and B in the vapour phase using the Dalton law of partial pressure as follows:}$ $y_A = \frac{P_A}{P_T} = \frac{200}{500} = \frac{2}{5} = 0.4$ $y_B = \frac{P_B}{P_T} = \frac{300}{500} = \frac{3}{5} = 0.6$ $\text{Or}$ $y_B = 1 - y_A$ $= 1 - 0.4$ $= 0.6$ $\text{Hence, option third is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}