If NaCl is doped with 10^(-3) mole % of ...

If NaCl is doped with `10^(-3)` mole % of `SrCl_(2)`, what is the concentration of the cation vacancies?

Text Solution

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Due to the addition of `SrCl_(2)` each `Sr^(2+)` ion replaces `2Na^(+)` ions, but occupies only one lattice point in place of two `Na^(+)` ion. This creates cation vacancy.
Number of moles of cation vacancies in 100 moles of NaCl is equal to `10^(-3)` mole.
Number of moles of cation vacancies in 1 mole of
`NaCl=(10^(-3))/(100)=10^(-5)`
Total number of cation vacancies
`=10^(-5)N_(A)=(10^(-5))xx (6.022xx 10^(23)"mol"^(-1))`
`=6.022 xx 10^(18)"mol"^(-1)`

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