Q76 / 200 · NEET 2007 · Chemistry
If $\mathrm{NaCl}$ is doped with $10^{-4} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the concentration of cation vacancies will be: $\left(\mathrm{N}_A=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)$
- A$6.02 \times 10^{16} \mathrm{~mol}^{-1}$
- Correct answerB$6.02 \times 10^{17} \mathrm{~mol}^{-1}$
- C$6.02 \times 10^{14} \mathrm{~mol}^{-1}$
- D$6.02 \times 10^{15} \mathrm{~mol}^{-1}$
Solution
Since each $\mathrm{Sr}^{++}$ion provides one cation vacancy, hence Concentration of cation vacancies $=$ mole $\%$ of $\mathrm{SrCl}_2$ added
$\begin{aligned}
& =10^{-4} \text { mole } \% \\
& =\frac{10^{-4}}{100} \times 6.023 \times 10^{23} \\
& =6.023 \times 10^{17} \mathrm{~mol}^{-1}
\end{aligned}$
Related Theory
Doping is the process of adding impurities to intrinsic semiconductors to alter their properties. When an intrinsic semiconductor is doped with Trivalent impurity it becomes a P-Type semiconductor. The $P$ stands for Positive, which means the semiconductor is rich in holes or Positive charged ions.
Hence, the correct answer is option: 'B'.