Question

The concentration of cation vacancies when $NaCl$ is doped with ${10}^{-3}$ mole percent of $SrC{l}_2$ is :

• A. $6.023\times {10}^{20}$
• B. $6.023\times {10}^{23}$
• C. $6.023\times {10}^{21}$
• D. $6.023\times {10}^{18}$

Answer 1

Answer: (D)

$6.023\times {10}^{18}$

Given, concentration of $SrC{l}_2={10}^{-3}$ mol percent

The moles of $SrC{l}_2$ are very negligible as compared to the total moles.

$1$ mol of NaCl is doped with $=\frac{{10}^{-3}}{100}$ $={10}^{-5}$ moles of $SrC{l}_2$

So, cation vacancies per mole of $NaCl$ $={10}^{-5}$ moles

$1$ mol $=6.022\times {10}^{23}$ particles

So, the cation vacancies per mole of $NaCl$ $={10}^{-5}\times 6.022\times {10}^{23}=6.02\times {10}^{18}$

So, the concentration of cation vacancies created by $SrC{l}_2$ is $6.022\times {10}^{18}$ per mol of $NaCl$.