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  • If NaCl is doped with 10–3 mol % of SrCl^2, what is the concentration of cation vacancies?

1.25     If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation vacancies?

 

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NaCl is doped with 10–3 mol % of SrCl_2

Concentration in % so that take a total of 100 mol of solution.

Moles of NaCl = 100 - moles of SrCl_2

Moles of SrCl_2 is very less , so we can neglect them.

Moles of NaCl =100

1 mole of NaCl is dipped with = 10^{-5}  mol of SrCl_2.

So cation vacanties per mole of NaCl =10^{-5} mol 

1mol=6.022\times 10^{23} particles 

So cation vacancies per mol of NaCl = 10-5 \times 6.022\times 10^{23}

                                                          =6.02\times 10^{18}

Hence, the concentration of cation vacancies.=6.02\times 10^{18}

Posted by

seema garhwal

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