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Current Question (ID: 18120)

Question:

$\text{Aqueous solutions of } 0.004 \text{ M Na}_2\text{SO}_4 \text{ and } 0.01 \text{ M Glucose are isotonic. The degree of dissociation of Na}_2\text{SO}_4 \text{ is-}$

Options:

1. $25\%$
2. $60\%$
3. $75\%$
4. $85\%$

Solution:

$\text{Hint: isotonic solutions have the same osmotic pressure}$ $\text{Step 1:}$ $\text{The osmotic pressure of two solutions are as follows:}$ $\pi_1 = C_1i_1 \text{ for } (0.004 \text{ M Na}_2\text{SO}_4)$ $\pi_2 = C_2i_2 \text{ for } (0.01 \text{ M Glucose}) \text{ for } i_2 = 1$ $C_1i_1 = C_2i_2$ $\text{Step 2:}$ $\text{Calculate the value of } i \text{ for Na}_2\text{SO}_4$ $i_1 = \frac{C_2}{C_1} \times 1 = \frac{0.01}{0.004 \times 1} = 2.5$ $\text{Step 3:}$ $\text{Na}_2\text{SO}_4 \rightleftharpoons 2 \text{Na}^+ + \text{SO}_4^{2-}$ $0.004 \quad \rightarrow \quad 0.004 - 0.004\alpha \quad \rightarrow \quad 0.008\alpha \quad \rightarrow \quad 0.004\alpha$ $i = \frac{0.004 - 0.004\alpha + 0.008\alpha + 0.004\alpha}{0.004}$ $i = 1 + 2\alpha$ $2.5 - 1 = 2\alpha$ $\alpha = 0.75$ $\alpha = 75\%$

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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
}