The Solid State - Result Question 36
####36. If $NaCl$ is doped with $10^{-4} mol %$ of $SrCl_2$, the concentration of cation vacancies will be
$(N_A=6.02 \times 10^{23} mol^{-1})$
[2007]
(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) $6.02 \times 10^{15} mol^{-1}$
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Solution:
- (b) For each $Sr^{2+}$ ion added, one $Na^{+}$ion is removed to maintain the electrical neutrality. Hence concentration of cation vacancies $=$ mole $%$ of $SrCl_2$ added $=10^{-4} mole %$
$=\frac{10^{-4}}{100} \times 6.023 \times 10^{23}=6.023 \times 10^{17}$
