Question

If NaCl is doped with ${10}^{-3}mol$ of $SrC{l}_2$, then what is the concentration of cation vacancies?

Answer 1

Given concentration of $SrC{l}_2={10}^{-3}mol\%$
Concentration is in percentage so that take total 100 mol of solution
Number of moles of $NaCl={100}^{-3}$ moles of $SrC{l}_2$
Moles of $SrC{l}_2$ is very negligible as compared to total moles. Percentage is always taken on100 so 1 mol of NaCl is dipped with = $\frac{{10}^{-3}}{100}$ moles of $SrC{l}_2$
$={10}^{–5}$ mol of $SrC{l}_2$
So cation vacancies per mole of $NaCl={10}^{–5}$ mol
1 mol $=6.022\times {10}^{23}$ particles
So, cation vacancies per mole of $NaCl={10}^{–5}\times 6.022\times {10}^{23}$
$=6.02\times {10}^{18}$
So that, the concentration of cation vacancies created by $SrC{l}_2$ is $6.022\times {10}^8$ per mol of NaCl.