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If NaCl is doped with 10^(-2) mol percen...

If NaCl is doped with `10^(-2)` mol percentage of strontium chloride , what is the concentration of cation vacancy ?

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Verified by Experts

Given concentration of ` Srcl_(20 = 10^(-3) mol %`
concentration is in percentage so that as take total 100 mol of solution .
No. of mol of NaCI = 100 - moles of ` SrCI_(2)`
Mol of ` SrCI_(2)`is very negligible as compare to total moles .
So No. of mol of NaCI = 100
1 mol of NaCI is dipped with
` = 10^(-3) // 100` moles of ` SrCI_(2)`
` 10^(-5) mol " of " SrCI_(2)`
So cation vacancies per mole of NaCI = `10^(-5)` mol
1 mole =` 6.022 xx 10^(23)` particles .
So cation vacancies per mol of NaCI
` 10^(-5) xx 6.022 xx 10^(23)`
` 6.02 xx 10^(18)`
So , that the concentration of cation vacancies created by ` SrCI_(2) is 6.022 xx 10^(18)` pre mol of NaCI.
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