The Solid State

Question

If NaCl is doped with 10^-4 mol% of SrCl₂ the concentration of cation vacancies will be
(N_A = 6.02 x 10²³ mol^-1):

6.02 x 10¹⁵ mol^-1

6.02 x 10¹⁶ mol^-1

6.02 x 10¹⁷

6.02 x 10¹⁴

Answer

6.02 x 10¹⁵ mol^-1

If NaCl is doped with 10^-4 mol% of SrCl₂,
2 Na⁺ ions doped by Sr²⁺ N_A = 6.02 x 10²³
The concentration of cation vacancies
= 6.02 x 10²³ x10^-8
= 6.02 x 10¹⁵ mol^-1

Sponsor Area

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Sponsor Area

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Sponsor Area

If NaCl is doped with 10^-4 mol% of SrCl₂ the concentration of cation vacancies will be
(N_A = 6.02 x 10²³ mol^-1):

6.02 x 10¹⁵ mol^-1

6.02 x 10¹⁶ mol^-1

6.02 x 10¹⁷

6.02 x 10¹⁴

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Mock Test Series

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Entrance Exams Preparation

If NaCl is doped with 10-4 mol% of SrCl2 the concentration of cation vacancies will be (NA = 6.02 x 1023 mol-1): 6.02 x 1015 mol-1 6.02 x 1016 mol-1 6.02 x 1017 6.02 x 1014

Some More Questions From The Solid State Chapter

Mock Test Series

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Entrance Exams Preparation

If NaCl is doped with 10^-4 mol% of SrCl₂ the concentration of cation vacancies will be
(N_A = 6.02 x 10²³ mol^-1):

6.02 x 10¹⁵ mol^-1

6.02 x 10¹⁶ mol^-1

6.02 x 10¹⁷

6.02 x 10¹⁴