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Question: Answered & Verified by Expert
\( 1 \) mole of \( \mathrm{NaCl} \) is doped with \( 10^{-5} \) mole of \( \mathrm{SrCl}_{2} . \) The number of cationic vacancies in the
crystal lattice will be:
ChemistrySolid StateKCETKCET 2019
Options:
  • A \( 6.022 \times 10^{23} \)
  • B \( 12.044 \times 10^{20} \)
  • C \( 6.022 \times 10^{18} \)
  • D \( 6.022 \times 10^{15} \)
Solution:
1620 Upvotes Verified Answer
The correct answer is: \( 6.022 \times 10^{18} \)
(C)
\( 1 \mathrm{~mol} \mathrm{SrCl}_{2} \) gives \( 1 \) cationic vacancy.
\( 10^{-5} \) moles of \( \mathrm{SrCl}_{2} \) gives \( 10^{-5} \) mole cationic vacancies.
\( \therefore \) The number of cationic vacancy in \( 1 \) moles of \( \mathrm{NaCl} \) when it is doped with \( 10^{-5} \) moles of \( \mathrm{SrCl}_{2} \) is \( 6.022 \times 10^{18} \)

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