1 text{mole} of NaCl is doped with 10^{-5} | vedclass

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(C) When $SrCl_{2}$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^{+}$ ions to maintain electrical neutrality. One $Sr^{2+}$ ion creates one

Vedclass · Accepted Answer

$6.022 	imes 10^{18}$

Vedclass · Answer

(C) When $SrCl_{2}$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^{+}$ ions to maintain electrical neutrality. One $Sr^{2+}$ ion creates one cationic vacancy. Therefore,$1 \ 	ext{mole}$ of $SrCl_{2}$ creates $1 \ 	ext{mole}$ of cationic vacancies. $10^{-5} \ 	ext{mole}$ of $SrCl_{2}$ will create $10^{-5} \ 	ext{mole}$ of cationic vacancies. Number of cationic vacancies $= 10^{-5} 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{18}$.

$1 \ \text{mole}$ of $NaCl$ is doped with $10^{-5} \ \text{mole}$ of $SrCl_{2}$. The number of cationic vacancies in the crystal lattice will be:

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