Question
If $$NaCl$$ is doped with $${10^{ - 4}}mol\,\,\% $$ of $$SrC{l_2},$$ the concentration of cation vacancies will be $$\left( {{N_A} = 6.023 \times {{10}^{23}}mo{l^{ - 1}}} \right)$$
A.
$$6.023 \times {10^{15}}mo{l^{ - 1}}$$
B.
$$6.023 \times {10^{16}}mo{l^{ - 1}}$$
C.
$$6.023 \times {10^{17}}mo{l^{ - 1}}$$
D.
$$6.023 \times {10^{14}}mo{l^{ - 1}}$$
Answer :
$$6.023 \times {10^{17}}mo{l^{ - 1}}$$
Solution :
Doping of $$NaCl$$ with $${10^{ - 4}}mol\,\% $$ of $$SrC{l_2}$$ means, $$100\,moles$$ of $$NaCl$$ are doped with $${10^{ - 4}}mol$$ of $$SrC{l_2}.$$
$$\therefore \,\,1\,mol$$ of $$NaCl$$ is doped with
$$SrC{l_2} = \frac{{{{10}^{ - 4}}}}{{100}} = {10^{ - 6}}\,mole$$
As each $$S{r^{2 + }}\,ion$$ introduces one cation vacancy.
$$\therefore $$ Concentration of cation vacancies
$$\eqalign{
& = {10^{ - 6}}mol/mol\,{\text{of}}\,NaCl \cr
& = {10^{ - 6}} \times 6.023 \times {10^{23}}mo{l^{ - 1}} \cr
& = 6.023 \times {10^{17}}mo{l^{ - 1}} \cr} $$