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Current Question (ID: 21420)

Question:

$2$ \text{ grams of non-volatile solute dissolves in } 200 \text{ grams each of two different solvents A and B to attain the same molality. If the ratio of } K_b \text{ of solvent A and B is } 1:8 \text{ and the ratio of elevation in boiling points is } x:y, \text{ then the value of } y \text{ is:}$ $1. \ 4$ $2. \ 6$ $3. \ 10$ $4. \ 8$

Options:

1. $4$
2. $6$
3. $10$
4. $8$

Solution:

$\text{Hint: The molality of two solutions is the same}$ $\text{Step 1:}$ $\text{The formula for the elevation of the boiling point is as follows:}$ $\Delta T_b = K_b \times m$ $\text{The molality of both solutions is the same. The equations are as follows:}$ $(\Delta T_b)_1 = (K_b)_1 \times m \quad \ldots(1)$ $(\Delta T_b)_2 = (K_b)_2 \times m \quad \ldots(2)$ $\text{Step 2:}$ $\text{Take the ratio of equations 1 and 2 as follows:}$ $\frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = \frac{(T_b)_I \times m}{(T_b)_{II} \times m} = \frac{1}{8} = \frac{x}{y}$ $\text{So } y = 8$

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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
}