If ${\rm{NaCl}}$ is doped with ${10^{ - 3}}$ mol % ${\rm{SrC}}{{\rm{l}}_2}$, then the concentration of cation vacancies will be

JEE CHEMISTRY MCQ: SOLID STATE

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Application +4 | -1

\mathrm{NaCl}

is doped with

{10^{ - 3}}

mol %

\mathrm{SrC}{\mathrm{l}_2}

, then the concentration of cation vacancies will be

View answer and explanation

Correct option: A

1 \times {10^{ - 3}}mol\%

Option A (

1 \times {10^{ - 3}}mol\%

) is correct because each SrCl2 introduces one cation vacancy.

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