Question

NaCl is doped with \(10^{-3}\) mole % of \(\mathrm{CaCl_2}\). The number of cationic vacancies in one mole of NaCl lattice is \[ (N = 6.02\times10^{23}\ \mathrm{mol^{-1}}) \]

• A. \(6.02\times10^{18}\)
• B. \(3.01\times10^{18}\)
• C. \(1.204\times10^{19}\)
• D. \(6.02\times10^{17}\)

Hint:

Doping NaCl with \(\mathrm{CaCl_2}\) produces one \(\mathrm{Na^+}\) vacancy for every \(\mathrm{Ca^{2+}}\) ion introduced.

Answer 1

The Correct Option is A

Step 1: Understand the defect. Each \(\mathrm{Ca^{2+}}\) ion replaces two \(\mathrm{Na^+}\) ions and creates \[ \boxed{\text{one cation vacancy}.} \]

Step 2: Calculate the number of CaCl\(_2\) molecules added. \[ 10^{-3}\text{ mole \%} = 10^{-3}\times\frac{1}{100} =10^{-5} \] Thus, moles of \(\mathrm{CaCl_2}\) added per mole of NaCl \[ =10^{-5}\ \text{mol} \] Hence, cation vacancies \[ =10^{-5}\times6.02\times10^{23} =6.02\times10^{18}. \] Therefore, \[ \boxed{6.02\times10^{18}} \] Hence, \[ \boxed{(A)} \] is the correct answer.