Question

If NaCl is doped with ${10}^{-3}$ mol percent of $SrC{l}_2$, what is the number of cation vacancies?

• A. $3.01\times {10}^{18}$ mol${}^{-1}$
• B. $6.02\times {10}^{18}$ mo$l^{-1}$
• C. $1.2\times {10}^{19}$ mol${}^{-1}$
• D. $18\times {10}^{18}$ mol${}^{-1}$

Answer 1

Answer: (B)

$6.02\times {10}^{18}$ mo$l^{-1}$

If NaCl is doped with ${10}^{-3}$ mol percent of $SrC{l}_2$, what is the number of cation vacancies is $6.02\times {10}^{18}$ mo$l^{-1}$.

$100$ moles of $NaCl$ contains ${10}^{-3}$ moles of $SrC{l}_2$

1 mole of NaCl will contain $6.02\times {10}^{23}\times \frac{{10}^{-3}}{100}=6.02\times {10}^{18}$ molecules of $SrC{l}_2$.

1 molecule of $SrC{l}_2$ will lead to one cation vacancy.

$6.02\times {10}^{18}$ molecules of $SrC{l}_2$ will give $6.02\times {10}^{18}$ cation vacancies