[Nitin_Naudiyal]

If NaCl is doped with 10^-4 mol % of Sr Cl₂, the concentration of cation vacancies will be
(NA = 6.02 x 10²³ mol^-)

Answer: 6.02 x 10¹⁷ mol^-1

The Solution given is as follows:

On doping NaCl by SrCl₂, one Sr²⁺ ion replaces two Na⁺ ion.
So, no. of moles of cation vacancy in 100ml NaCl = 10^-4
no of moles of cation vacancy in 1 mole NaCl = 10^-4/100 = 10^-6
Thus, total cation vacancy = 10^-6 x N_A = 10^-6 x 6.022 x 10²³ = 6.022 x 10¹⁷

I have the Following questions:
How does one Sr²⁺ ion replaces two Na⁺ ion?
How is this data usefull in knowing the concentration of cation vacancies?

Thanking in Advance

[UG]

I have the Following questions:
How does one Sr²⁺ ion replaces two Na⁺ ion?
How is this data usefull in knowing the concentration of cation vacancies?

The crystal must maintain a neutral charge so a Sr²⁺ ion will displace 2 Na⁺ ions. If you know how many strontium ions there are, you know how many Na⁺ ions have been displaced and therefore the amount of vacancies in your crystal. For example if I doped the crystal with 20 strontium ions, 40 sodium ions will be displaced so I will have 20 vacancies or 'holes' left in the crystal.

[AWK]

So, no. of moles of cation vacancy in 100ml NaCl = 10-4

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