Question

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

Answer 1

Given Conetration of$SrC{l}_2={10}^{-3}$ mol%

Concentration is in percentage so that take total 100 mol of solution
Number of moles of $NaCl=100–molesofSrC{l}_2$
Moles of $SrC{l}_2$is very negligible as compare to total moles so
Number of moles of $NaCl=100$
1 mol of NaCl is dipped with $=10-\frac{3}{100}$moles of$SrC{l}_2={10}^{-5}$ mol of $SrC{l}_2$
So cation vacancies per mole of $NaCl={10}^{-5}mol$
$1mol=6.022\times {10}^{23}$ 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 $NaCl$.