1.

NaCl is doped with 2xx10^(-3) mole % of SrCl_(2). The concentration of cation vacancies is

Answer»

`12.04xx10^(20)` per mole
`3.01xx10^(18)` per mole
`6.02xx10^(18)` per mole
`12.04xx10^(18)` per mole

Solution :The ADDITION of one `Sr^(2+)` REPLACES `2Na^(+)` and one cationic vacancy is created
No. of cationic vacancy `=2xx10^(-3)` mole % of NaCl
`=(2xx10^(-3))/(100) mol^(-1)` of NaCl
`=2xx10^(-5)xx6.02xx10^(23)mol^(-1)`
`=12.04xx10^(18)mol^(-1)` of NaCl
`=2xx10^(-5)xx6.02xx10^(23)mol^(-1)`
`=12.04xx10^(18)mol^(-1)` of NaCl.


Discussion

No Comment Found

Related InterviewSolutions