In 1g of KBr, `10^(-5)` mole percent `SrBr_2` is doped. Find number of cationic vacancies

KBr is doped with 10^(-5) mole percent of SrBr_(2) . The number of cationic vacancies in 1 g of KBr crystal is ________ 10^(14) . (Round off to the Nearest Integer). [Atomic Mass : K : 39.1 u, Br : 79.9 u N_(A)=6.023 xx 10^(23) ]

If NaCI is droped with 10^(-3) mole of SrCI_(2) then number of cationic vacancies is

If 1 mole of NaCl is doped with 10^(-3) mole of SrCl_(2) . What is the number of cationic vacancies per mole of NaCl ?

If NaCI is droped with 10^(-4) "mole"% of SrCI_(2) then number of cationic vacancies will be

If NaCl is doped with 10^(-3) mol percent of SrCI_(2) , what is the concentration of cation vacancy?

When NaCl is dopped with 10^(-5) "mole % of" SrCl_(2) , what is the no. of cationic vacanies?

When NaCl is dopped with 10^(-8) mole % of SrCl . What is the no. of cationic vacancies?