NaCl is doped with 2×10^–2 mole% of SrCl₂ then the number of cation vacancies per mole

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$12.046 \times 10^{19}$

$12.046 \times 10^{21}$

$6.023 \times 10^{21}$

$12.046 \times 10^{18}$

answer is B.

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Detailed Solution

Given concentration of SrCl₂=2 $\times$ 10⁻² mol%
concentration is in percentage so that take total 100 mol of solution.
No. of mol. of NaCl = 100- moles of SrCl₂
Mol. of SrCl₂ is very negligible as compare to total moles.
So No. of mole of NaCl = 100
1 mol of NaCl is dipped with
=2x10⁻²/100 moles of SrCl₂
=2x10⁻⁴ mole of SrCl₂
So cation vacancies per mole of NaCl = 2x10⁻⁴ mol
1 mole =6.023×10²³ particles .
So cation vacancies per mole of NaCl = 2x10⁻⁴ mol
= 2x10⁻⁴×6.023×10²³
=12.046x10¹⁹
So, that the concentration of cation vacancies created by SrCl₂ is 12.046x10¹⁹ per mole of NaCl.