Question

If MgCl2 is doped with 10^(-4) mol% of AlCl3 .Find Cation Vacancy.

Answer 1

Answer:

IF nacl is doped with 10 to the power minus 4 mole percent SRCL to the concentration of cation vacancies will be

If NaCl is doped with 10^-4 mol % of SrCl2, theconcentration of cation vacancies will be (NA = 6.02 x 10^23 mol^-1)

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Answer 2

Answer:

The total number of cation vacancies formed on doping of MgCl₂ crystal with AlCl₃ is 3.0115×10¹⁷.

Explanation:

Number of moles of AlCl₃ in 1 mole of MgCl₂ $=\frac{10^{-4}}{100} =10^{-6}moles$

When MgCl₂ crystal is doped with AlCl₃, to keep the crystal electrically neutral, three $Mg^{2+}$ replaced by two $Al^{3+}$ and one cation vacancy is created.

Therefore 2 moles of AlCl₃  will form = $1mol$ of cation vacancies

$10^{-6}moles$ AlCl₃ will form cation vacancies $=\frac{10^{-6}}{2}=0.5*10^{-6}mol$

Total number of cation vacancies in  MgCl₂ crystal will be: $=0.5*10^{-6}*6.023*10^{23}$

$=3.0115*10^{17}$ cation vacancies