The concentration of cation vacancies per mole when NaCl is doped with `10^(-3)` mole % of `SrCl_(2)`

What is the concentration of cation vacancies if NaCl is doped with 10^(-3) mol percent of MgCl_(2) ?

Calculate the concentration of cation vaccancies if Nacl dopped with 10^-3 mole % of SrCl

Read the following passage and answer the questions at the end of it. Conductivity of silicon increases if it is doped with certain other elements. Doping means introduction of small amount of impurities like phosphorus, arsenic or boron in the pure crystal. In pure silicon four valencies are used in bonding with other four adjacent silicon atoms. When a silicon crystal is doped with a group - 15 element (with five valence electrons) such as P, AsSb or Bi, the structure of the crystal lattice remains unchanged. Out of the five valence electrons of group - 15 doped element, four electrons are used in normal covalent bonding with silicon and fifth electron is delocalised and thus conducts electricity Doping a silicon crystal with a group-13 element (with three valence electron) such as B, Al, Ga or In produces a semiconductor with three electrons in dopant. The place where fourth electron is missing is called as electron vacancy or hole. If NaCl is doped with 10^(-3)" mol % " SrCl_(2) then concentration of cation vacancies is

If NaCl is doped with 10^(4) mol % of SrCl_(2) the concentration of cation vacancies will be (N_(A) = 6.02 xx 10^(23) mol^(-1))

What type of defect is produced when NaCl is doped with SrCl_(2) ?

If NaCl is doped with 10^(-3) mole percent SrCl_(2) , what will be the concentration of cation vacancies ? ( N_(A) = 6.02 xx 10^(23) mol^(-1) )

If NaCl is doped with 10^(-4) mol % of SrCl_(2) , the concentration of cation vacancies will be ( N_(A) = 6.022 xx 10^(23) mol^(-1))