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Current Question (ID: 21394)

Question:

$\text{The vapour pressure of pure A is 50 mm of Hg and the vapour pressure of pure B is 100 mm of Hg in a mixture of liquid A and B mole fraction of A is 0.3, then the mole fraction of B in the vapour phase is } \frac{x}{17}. \text{ The value of } x \text{ is:}$

Options:

1. $14$
2. $17$
3. $20$
4. $23$

Solution:

$\text{Hint: } P_{\text{Total}} = P_A^o \times X_A + P_B^o \times X_B$ $P_{\text{Total}} = P_A^o \times X_A + P_B^o \times X_B$ $= (50)0.3 + (100)0.7$ $= 15 + 70$ $= 85 \text{ mm of Hg}$ $P_B = (P_{\text{Total}}) Y_B$ $\Rightarrow 70 = (85)Y_B$ $\Rightarrow Y_B = \frac{70}{85} = \frac{14}{17}$ $\text{The value of } x \text{ is 14.}$

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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
}