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\( 1 \) 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:
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crystal lattice will be:
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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 \( 6.022 \times 10^{18} \)
\( 1 \mathrm{~mol} \mathrm{SrCl}_{2} \) 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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