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Current Question (ID: 18082)

Question:

$\text{Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are } 50.71 \text{ mm Hg and } 32.06 \text{ mm Hg respectively.}$ $\text{The mole fraction of benzene in vapour phase, if 80 g of benzene is mixed with 100 g of toluene, would be:}$

Options:

1. 0.41
2. 0.68
3. 0.72
4. 0.59

Solution:

$\text{Hint: The relative lowering of vapor pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Calculate the mole of toluene and benzene as follows:}$ $\text{Number of moles in 80 g of benzene} = \frac{80g}{78 \text{ g mol}^{-1}} = 1.026 \text{ mol}$ $\text{Number of moles in 100 g of toluene} = \frac{100}{92 \text{ g mol}^{-1}} = 1.087 \text{ mol}$ $\text{Mole fraction of benzene, } x_B = 0.486; \text{ Mole fraction of toluene, } x_T = 0.514$ $\text{Step 2:}$ $\text{Partial vapour pressure of benzene; } p_B = x_B \times p_B^\circ = 0.486 \times 50.71 = 24.645 \text{ mm Hg}$ $\text{Partial vapour pressure of toluene; } p_T = x_T \times p_T^\circ = 0.514 \times 32.06 = 16.479 \text{ mm Hg}$ $\text{So, the mole fraction of Benzene in vapour phase is; } \frac{p_B}{p_B + p_T} = \frac{24.645}{24.645 + 16.479} = 0.59$

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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
}