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If NaCl is doped with 10^(-4)mol%of SrC...

If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`

A

`6.02 xx 10^(16) mol^(-1)`

B

`6.02 xx 10^(17) mol^(-1)`

C

`6.02 xx 10^(14) mol^(-1)`

D

`6.02 xx 10^(15) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
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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Knowledge Check

  • If NaCl is doped with 10^(-4) mol% "of" SrCl_(2) the concentration of cation vacancies will be (N_(A)=6.02times10^(23)mol^(-1))

    A
    `6.02times10^(23) mol^(-1)`
    B
    `6.02times10^(15) mol^(-1)`
    C
    `6.02times10^(16) mol^(-1)`
    D
    `6.02times10^(17) mol^(-1)`

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

    A
    ` 6.02 xx 10^(14) mol^(-1)`
    B
    `6.02 xx 10^(15) mol^(-1)`
    C
    ` 6.02 xx 10^(16) mol^(-1)`
    D
    ` 6.02 xx 10^(17) mol^(-1) `

  • NaCl is doped with 2xx10^(-3) mol % SrCl_2 , the concentration of cation vacancies is

    A
    `3.01xx10^(18) "mol"^(-1)`
    B
    `12.04 xx10^(18) "mol"^(-1)`
    C
    `6.02xx10^(18) "mol" ^(-1)`
    D
    `12.04 xx10^(20) "mol"^(-1) `

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    If NaCI is doped with 10^(-3)"mol"%SrCI_(2) then the concentration of cation vacancies will be:

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