If NaCl crystals are doped with `2xx10^(-3)` mol per cent of `SrCl_(2)`, calculate the cation vacancies per mole.

Options

A. `1.204xx10^(20)`
B. `12.04xx10^(20)`
C. `12.04xx10^(19)`
D. `1.204xx10^(19)`

Answer

D

Explanation

One `Sr^(2+)` will replace `2Na^(+)` which will result in one cation vacancy.
Consider 1 mole doped NaCl, `Sr^(2+)=6.02xx10^(23)xx(2xx10^(-3))/(100)`
Cation vacancy=`12.04xx10^(18)=1.204xx10^(19)`.