Saved Bookmarks
| 1. |
If NaCl is doped with 10^(-3) mol % of SrCl_2, what is the concentration of cation vacancies ? |
|
Answer» SOLUTION :Let moles of NaCl in the crystal = 100 Moles of `SrCl_2` doped = `10^(-3)` `therefore` No. of `NA^(+)` ions replaced by each `SR^(2+)` ions = 1 No. of CATIONIC vacancies formed by `Sr^(2+)` ions = 1 `therefore` Number of cationic vacancies = `10^(-3)` formed by `10^(-3)` mol `SrCl_2` `therefore` Vacancy per mole of `NaCl =(10^(-3))/(100) = 10^(-5)` `therefore` Number of vacancies per mole of NaCl ` = 6.022 xx 10^23 X 10^(-5)` ` = 6.022 xx 10^18`. |
|