Import Question JSON

Current Question (ID: 18205)

Question:

$\text{The specific conductivity of a solution depends upon:}$ $1. \text{Number of ions as well as mobility of ions.}$ $2. \text{Number of ions per cc or (cm}^3\text{) of solution.}$ $3. \text{Number of ions per cc or (cm}^3\text{) as well as mobilities of ions.}$ $4. \text{Mobilities of ions.}$

Options:

1. $\text{Number of ions as well as mobility of ions.}$
2. $\text{Number of ions per cc or (cm}^3\text{) of solution.}$
3. $\text{Number of ions per cc or (cm}^3\text{) as well as mobilities of ions.}$
4. $\text{Mobilities of ions.}$

Solution:

$\text{HINT: Specific conductance is the conductance per c.c. solution}$ $\text{Explanation:}$ $\text{STEP 1: Conductivity = conductance } \times \text{ cell constant}$ $K = \text{conductivity, } G = \text{conductance, } l = \text{length and } A = \text{area of cross-section}$ $\text{STEP 2:}$ $\text{The conductance of a material depends on the:}$ $1. \text{Nature of the material,}$ $2. \text{Number of valence electrons for material}$ $3. \text{Temperature.}$ $\text{STEP 3:}$ $\text{Conductivity depends on:}$ $1. \text{Number of ions in 1 cc should be high.}$ $2. \text{Those ions must have higher mobility.}$ $\text{Hence, option 3 is the best-suited answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}