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If NaCl is doped with $10^{-4}$ mol % of $SrCl_2$, the concentration of cation vacancies will be
($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)
($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)
A
$6.02 \times 10^{16}$ mol$^{-1}$
B
$6.02 \times 10^{17}$ mol$^{-1}$
C
$6.02 \times 10^{14}$ mol$^{-1}$
D
$6.02 \times 10^{15}$ mol$^{-1}$
Detailed Solution
When $SrCl_2$ is added to NaCl, each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.
It occupies the site of one $Na^+$ ion and the other site remains vacant, so each $Sr^{2+}$ ion provides one cation vacancy.
Hence the concentration of cation vacancies = mole percent of $SrCl_2$ added $= 10^{-4}$ mole percent
That is, 100 moles of NaCl contain $10^{-4}$ mole of cation vacancies.
Cation vacancies per mole of NaCl $= \dfrac{10^{-4}}{100} = 10^{-6}$ mole
Number of cation vacancies $= \dfrac{10^{-4}}{100} \times 6.02 \times 10^{23}$
$= 6.02 \times 10^{17}$ mol$^{-1}$
It occupies the site of one $Na^+$ ion and the other site remains vacant, so each $Sr^{2+}$ ion provides one cation vacancy.
Hence the concentration of cation vacancies = mole percent of $SrCl_2$ added $= 10^{-4}$ mole percent
That is, 100 moles of NaCl contain $10^{-4}$ mole of cation vacancies.
Cation vacancies per mole of NaCl $= \dfrac{10^{-4}}{100} = 10^{-6}$ mole
Number of cation vacancies $= \dfrac{10^{-4}}{100} \times 6.02 \times 10^{23}$
$= 6.02 \times 10^{17}$ mol$^{-1}$