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KCET · Chemistry · Solid State

\( 1 \) 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:

  1. A \( 6.022 \times 10^{23} \)
  2. B \( 12.044 \times 10^{20} \)
  3. C \( 6.022 \times 10^{18} \)
  4. D \( 6.022 \times 10^{15} \)
Verified Solution

Answer & Solution

Correct Answer

(C) \( 6.022 \times 10^{18} \)

Step-by-step Solution

Detailed explanation

(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 \( 6.022 \times 10^{18} \)