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Current Question (ID: 18197)

Question:
$\text{The molar conductivity of a } 0.5 \text{ mol/dm}^3 \text{ solution of } \text{AgNO}_3 \text{ with electrolytic conductivity of } 5.76 \times 10^{-3} \text{ S cm}^{-1} \text{ at } 298 \text{ K is:}$
Options:
  • 1. $11.5 \text{ S cm}^2/\text{mol}$
  • 2. $21.5 \text{ S cm}^2/\text{mol}$
  • 3. $31.5 \text{ S cm}^2/\text{mol}$
  • 4. $41.5 \text{ S cm}^2/\text{mol}$
Solution:
$\text{HINT: } \Lambda_m = \frac{K \times 1000}{M}$ $\text{STEP 1:}$ $\text{Given, concentration of solution, } M = 0.5 \text{ mol/dm}^3$ $\text{Electrolytic conductivity, } K = 5.76 \times 10^{-3} \text{ S cm}^{-1}, T = 298 \text{ K}$ $1 \text{ dm}^3 = 1 \text{ litre}$ $\text{STEP 2: Molar conductivity,}$ $\Lambda_m = \frac{K \times 1000}{M}$ $= \frac{5.76 \times 10^{-3} \times 1000}{0.5}$ $= 11.52 \text{ S cm}^2/\text{mol}$

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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
}