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Current Question (ID: 21407)

Question:

$1 \text{ g of a non-volatile non-electrolyte solute is dissolved in } 100 \text{ g of two different solvents A and B whose ebullioscopic constants are in the ratio of } 1 : 5. \text{ The ratio of the elevation in their boiling points, } \frac{\Delta T_b(A)}{\Delta T_b(B)}, \text{ is:}$

Options:

1. $5:1$
2. $1:2$
3. $10:1$
4. $1:5$

Solution:

$\text{Hint: The formula of elevation in boiling point is } \Delta T_b = k_b m$ $\text{Step 1:}$ $\text{The two solutions are formed by dissolving 1 g of a non-volatile non-electrolyte solute in 100 g of two different solvents A and B.}$ $\text{The amount of solute in both the solution is same and amount of solvent is also same. Hence, molality of both the solution is same.}$ $\text{Step 2:}$ $\text{The formula of elevation in boiling point for solution I and solution II are as follows:}$ $\Delta T_b(I) = k_{b(A)} m_I \quad \ldots (1)$ $\Delta T_b(II) = k_{b(B)} m_{II} \quad \ldots (2)$ $m_I = m_{II}. \text{ The ratio of } k_{b(A)} : k_{b(B)} \text{ is } 1 : 5.$ $\text{Step 3:}$ $\text{Calculate the ratio of the elevation in their boiling points by dividing equation 1 and 2 as follows:}$ $\frac{\Delta T_b(I)}{\Delta T_b(II)} = \frac{k_{b(A)} m_I}{k_{b(B)} m_{II}}$ $\frac{\Delta T_b(I)}{\Delta T_b(II)} = \frac{1}{5}$

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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
}