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Current Question (ID: 18079)

Question:

$\text{Total vapour pressure of a mixture of 1 mol A } (p_A^\circ = 150 \text{ torr}) \text{ and 2 mol B } (p_B^\circ = 240 \text{ torr}) \text{ is 200 torr. In this case:-}$ $1. \text{ There is a positive deviation from Raoult's law}$ $2. \text{ There is a negative deviation from Raoult's law}$ $3. \text{ There is no deviation from Raoult's law}$ $4. \text{ None of these}$

Options:

1. $\text{There is a positive deviation from Raoult's law}$
2. $\text{There is a negative deviation from Raoult's law}$
3. $\text{There is no deviation from Raoult's law}$
4. $\text{None of these}$

Solution:

$\text{HINT: If the vapour pressure of a mixture is lower than expected from Raoult's law, there is said to be a negative deviation.}$ $\text{Step 1:}$ $\text{Calculate the total pressure as per the Raoult's Law as follows:}$ $p = p_A^\circ x_A + p_B^\circ x_B$ $p = 150 \left( \frac{1}{3} \right) + 240 \left( \frac{2}{3} \right)$ $p = 210 \text{ torr}$ $\text{Step 2:}$ $p_{\text{exp}} < p_{\text{cal}} ; \text{ Negative deviation from Raoult's law}$

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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
}