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))

Answer»

`6.02 xx 10^(18) MOL^(-1)`
`6.02 xx 10^(17) mol^(-1)`
`6.02 xx 10^(14) mol^(-1)`
`6.02 xx 10^(15) mol^(-1)`

Solution :ONE cation of `Sr^(2+)` would create one cation vacancy in NaCl. Therefore , the number of cation vacancies created in the LATTICE of NaCl is equal to the number of divalent `Sr^(2+)` ions added.
No. of moles of cationic vacancies ` = ( 10^(-4))/( 10^(2)) = 10^(-6)` mol
No. of cation vacancies
` = 10^(-6) xx 6.02 xx 10^(23) = 6.02 xx 10^(17)`


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