Q38 / 180 · KCET 2019 · Chemistry
\( 1 \) 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:
crystal lattice will be:
- A\( 6.022 \times 10^{23} \)
- B\( 12.044 \times 10^{20} \)
- Correct answerC\( 6.022 \times 10^{18} \)
- D\( 6.022 \times 10^{15} \)
Solution
(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 \( 6.022 \times 10^{18} \)
Hence, the correct answer is option: 'C'.