This chapter belongs to a previous NCERT/CBSE curriculum and is not part of the current 2026-27 chapter list. Removed in 2022-23.
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter The Solid State. This page offers a step-by-step solution to the specific question from Exercise 2, Question 25:
If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 25: If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?....
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