Question

NaCl is doped with ${10}^{-4}$ mol % $SrC{l}_2$, the concentration of cation vacancies is:
[Hint : Concentration of vacancies = $\frac{{10}^{-4}}{100}{N}_A$ ]

• A. $6.02\times {10}^{15}mo{l}^{-1}$
• B. $6.02\times {10}^{16}mo{l}^{-1}$
• C. $6.02\times {10}^{17}mo{l}^{-1}$
• D. $6.02\times {10}^{14}mo{l}^{-1}$

Answer 1

Answer: (C)

$6.02\times {10}^{17}mo{l}^{-1}$

It is given that NaCl is doped with ${10}^{-4}$ mol % of $SrC{l}_2$.

This means that 100 mol of NaCl is doped with ${10}^{-4}$ mol % of $SrC{l}_2$

Therefore, 1 mol of NaCl is doped with $\frac{{10}^{-4}}{100}$ mol of $SrC{l}_2$

=${10}^{-6}$

Cation vacancies produced by one ${Sr}^{2+}$ ion = 1

concentration of the cation vacancies produced by ${10}^{-6}$ mole of $SrC{l}_2$

$={10}^{-6}\times 6.022\times {10}^{23}$

$=6.022\times {10}^{17}$ per mole