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Current Question (ID: 18080)

Question:

$\text{The equal weight of a solute is dissolved in an equal weight of two solvents A and B to form a very dilute solution.}$ $\text{The relative lowering of vapour pressure for solution B has twice the relative lowering of vapour pressure for solution A.}$ $\text{If } M_A \text{ and } M_B \text{ are the molecular weights of solvents A and B respectively, then:}$

Options:

1. $M_A = M_B$
2. $M_B = 2M_A$
3. $M_A = 4M_B$
4. $M_A = 2M_B$

Solution:

$\text{HINT: Use general expression of relative lowering of vapour pressure.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{For A:}$ $\frac{P^0 - P}{P^0} = X_1 \text{ (mole fraction)}$ $= \frac{W/M_2}{W_1/M_A}$ $= \frac{W}{M_2} \times \frac{M_A}{W_1}$ $\text{For B:}$ $\frac{P^0 - P}{P^0} = 2X_1 \text{ (mole fraction)}$ $\text{Step 2:}$ $X_2 = 2 \times \frac{W}{M_2} \times \frac{M_A}{W_1}$ $\frac{W}{M_2} \times \frac{M_B}{W_1} = 2 \times \frac{W}{M_2} \times \frac{M_A}{W_1}$ $M_B = 2M_A$

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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
}