Question

if NaCl is doped with 10 ^-3 mol% of SrCl2, what is the concentration of cation vacancies?

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Answer 1

Given Conetration of SrCl2 = 10−3 mol%
Concentration is in percentage so that take total 100 mol of solution
Number of moles of NaCl = 100 – moles of SrCl2
Moles of SrCl2is very negligible as compare to total moles so
Number of moles of NaCl = 100
1 mol of NaCl is dipped with = 10−3/100 moles of SrCl2
= 10–5 mol of SrCl2
So cation vacancies per mole of NaCl =10–5 mol
1 mol = 6.022 x1023 particles
So
So cation vacancies per mole of NaCl = 10–5 x 6.022 x1023
= 6.02 x1018
So that, the concentration of cation vacancies created by SrCl2is 6.022 × 108 per mol of NaCl.


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Answer 2

Answer:

The number of cation vacancies created in NaCl crystal by doping of  SrCl₂ is equal to 6.023 × 10¹⁸.

Explanation:

When NaCl is  being doped with SrCl₂. The two Na⁺ ions are escaped from the crystal and one Sr²⁺ occupy one vacancy to keep it neutral. In the meantime one vacancy is created.

Therefore one Sr²⁺ creates one vacancy in the crystal lattice.

100 moles of NaCl  doped $=0.001moles$ of SrCl₂

1 mole of NaCl doped $=\frac{0.001}{100}=10^{-5}$ moles of SrCl₂

$10^{-5}moles$ of SrCl₂  $=10^{-5}moles$ of cation vacancies

                               $=10^{-5}*6.023*10^{23}=6.023*10^{18}$ cation vacancies

Therefore 10⁻³ mol% of SrCl₂ creates 6.023×10¹⁸ cation vacancies in a NaCl crystal lattice.