Question

If $NaCl$ is doped with $10^{-4}\, mol \, \% $ of $SrCl_2,$ the concentration of cation vacancies will be $(N_A = 6 . 0 2 3 \times 10^{23}\,mol^{-1})$

• A. $6 . 0 2 3 \times 10^{15}\,mol^{-1}$
• B. $6 . 0 2 3 \times 10^{16}\,mol^{-1}$
• C. $6 . 0 2 3 \times 10^{17}\,mol^{-1}$
• D. $6 . 0 2 3 \times 10^{14}\,mol^{-1}$

Answer 1

The Correct Option is C

Doping of $NaCl$ with $10^{-4} mol \% of \, SrCl_2$ means, $100$ moles of $NaCl$ are doped with $10^{-4} mol\, of\, SrCl_2$ $\therefore$ $1$ mol of $NaCl$ is doped with $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, SrCl_2=\frac{10^{-4}}{100}=10^{-6} mole$ As each $Sr^{2+}$ ion introduces one cation vacancy. $\therefore$ Concentration of cation vacancies $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10^{-6}$ mol of $NaCl$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10^{-6}\times6.023 \times10^{23}mol^{-1}$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =6.023\times10^{17} mol^{-1}$