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Question
If Nacl is doped with 10^ -3 mol % srcl2 ,what is the concentration of cation vacancies?
If Nacl is doped with 10^ -3 mol % srcl2 ,what is the concentration of cation vacancies?
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Solution
Answer: 6.02 x 1018 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 = 10-3
no of moles of cation vacancy in 1 mole NaCl = 10-3/100 = 10-5
Thus, total cation vacancy
= 10-5 x NA
= 10-5x 6.022 x 1023
= 6.022 x 1018
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 = 10-3
no of moles of cation vacancy in 1 mole NaCl = 10-3/100 = 10-5
Thus, total cation vacancy
= 10-5 x NA
= 10-5x 6.022 x 1023
= 6.022 x 1018
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