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Current Question (ID: 17943)

Question:

$\text{If NaCl is doped with } 10^{-5} \text{ mol\% of } \text{SrCl}_2, \text{ the number of cation vacancies will be -}$

Options:

1. $6.02 \times 10^{16}$
2. $6.02 \times 10^{18}$
3. $6.02 \times 10^{20}$
4. $2 \times 6.02 \times 10^{23}$

Solution:

$\text{If NaCl is doped with SrCl}_2 \text{ one cation vacancy will be produced by one Sr}^{2+} \text{ cation.}$ $\text{STEP 1:}$ $\text{The given data is as follows:}$ $\text{It is given that NaCl is doped with } 10^{-5} \text{ mol \% of SrCl}_2.$ $\text{This means that 100 mol of NaCl is doped with } 10^{-5} \text{ mol of SrCl}_2.$ $\text{Therefore, 1 mol of NaCl is doped with } \frac{10^{-5}}{100} \text{ of SrCl}_2 = 10^{-7} \text{ mol of SrCl}_2.$ $\text{STEP 2: Calculate the concentration of cation vacancies.}$ $\text{Cation vacancies produced by one Sr}^{2+} \text{ ion} = 1$ $\text{The concentration of the cation vacancies produced by } 10^{-7} \text{ mol of Sr}^{2+}$ $\text{ions} = 10^{-7} \times 6.023 \times 10^{23} = 6.023 \times 10^{16}$ $\text{Hence, the concentration of cation vacancies created by SrCl}_2 \text{ is } 6.022 \times 10^{16} \text{ per mol of NaCl.}$

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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
}