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Current Question (ID: 18055)

Question:

$\text{The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with an equimolar solution of benzene and toluene is -}$

Options:

1. $0.50$
2. $0.6$
3. $0.27$
4. $0.73$

Solution:

$\text{Hint: Vapour pressure } \propto \text{ mole fraction}$ $\text{Step 1:}$ $\text{For equimolar solutions, } X_B = X_T = 0.5$ $P_B = X_B \times P^0_B = 0.5 \times 160 = 80 \text{ mm}$ $P_T = X_T \times P^0_T = 0.5 \times 60 = 30 \text{ mm}$ $P_{\text{Total}} = 80 + 30 = 110 \text{ mm}$ $\text{Step 2:}$ $\text{As per Dalton's law of partial pressures}$ $P_{\text{partial pressure}} = P_{\text{total}} X$ $\text{Mole fraction of toluene in the vapour phase} = \frac{30}{110} = 0.27$

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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
}