If NaCl is doped with 10-3mol % of SrCl2 | Class 12 Chemistry Chapter The Solid State, The Solid State NCERT Solutions

Q25.

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

1 Cation of Sr2+ will create 1 cation vacancy in Nacl.Therefore the number of cation vacancies created in the lattice of Nacl is equal to the number of divalent Sr2+ ions added.

Now the concentration of cation vacancy on being doped with 10-3 mol% of Srcl2.

= 10-3 mol% = 10-3 / 100 = 10-5 mol\

Also number of Sr2+ ion in 10-5 mol = 10-5 x 6.023 x 1023 = 6.023 x 1018

Therefore number of cation vacancies = 6.023 x 1018

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What is the correct answer to: If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies??

1 Cation of Sr2+ will create 1 cation vacancy in Nacl.Therefore the number of cation vacancies created in the lattice of Nacl is equal to the number of divalent Sr2+ ions added.

Now the concentration of cation vacancy...

How do you solve If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies? step by step?

Step-by-step explanation:
• 1 Cation of Sr2+ will create 1 cation vacancy in Nacl
• Therefore the number of cation vacancies created in the lattice of Nacl is equal to the number of divalent Sr2+ ions added


• Now the concentration of cation vacancy on being doped with 10-3 mol% of Srcl2

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

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