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Current Question (ID: 18104)

Question:

$\text{The freezing point depression constant for water is } 1.86 \, ^\circ\text{C m}^{-1}. \text{ If } 5.00 \, \text{g Na}_2\text{SO}_4 \text{ is dissolved in } 45.0 \, \text{g H}_2\text{O}, \text{ the freezing point is changed by } -3.82 \, ^\circ\text{C}. \text{ The Van't Hoff factor for Na}_2\text{SO}_4 \text{ is:}$

Options:

1. $2.63$
2. $3.11$
3. $0.381$
4. $2.05$

Solution:

$\text{Hint: } \Delta T_f = i \times K_f \cdot m$ $\text{Step 1:}$ $\text{The formula of depression in the freezing point is as follows:}$ $\Delta T_f = i \times K_f \times \frac{w_B \times 1000}{m_B \times w_A}$ $\text{Step 2:}$ $\text{Calculate the value of } i \text{ as follows:}$ $i = \frac{\Delta T \times m_B \times w_A}{K_f \times w_B \times 1000}$ $= \frac{3.82 \times 142 \times 45}{1.86 \times 5 \times 1000} = 2.63$

Import JSON File

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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
}