AP EAMCET · Chemistry · Solid State
If AgCl is doped with \(1 \times 10^{-4}\) mole percent of \(\mathrm{CdCl}_2\), the number of cation vacancies ( \(\mathrm{in}\) \(\mathrm{mol}^{-1}\) ) is
- A \(6.023 \times 10^{19}\)
- B \(6.023 \times 10^{21}\)
- C \(6.023 \times 10^{17}\)
- D \(6.023 \times 10^{23}\)
Answer & Solution
Correct Answer
(C) \(6.023 \times 10^{17}\)
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