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If NaCl is doped with 10–2 mol% of SrCl2, which of the following option shows concentration of cation vacancies?
- a)10–4 mol–1
- b)6.023 × 1019 mol–1
- c)6.023 × 1021 mol–1
- d)106 mol–1
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If NaCl is doped with 10–2mol% of SrCl2, which of the following ...
Doping of NaCl with 10–2 mol% SrCl2 means that 100 mol of NaCl are doped with 10–2 mol of SrCl2.
100 mol of NaCl doped with = 10–2 mol SrCl2
1 mol of NaCl doped with
100 mol of NaCl doped with = 10–2 mol SrCl2
1 mol of NaCl doped with
Each 
ion introduces one cation vacancy,
therefore, cation vacancy: 10–4 mol/mol of NaCl
= 10–4 × 6.023 × 1023 mol–1
= 6.02 × 1019 mol–1 of NaCl
Hence, option B is correct.
ion introduces one cation vacancy,therefore, cation vacancy: 10–4 mol/mol of NaCl
= 10–4 × 6.023 × 1023 mol–1
= 6.02 × 1019 mol–1 of NaCl
Hence, option B is correct.
Most Upvoted Answer
If NaCl is doped with 10–2mol% of SrCl2, which of the following ...
Understanding the Problem
When NaCl is doped with SrCl2, the Sr2+ ions replace some Na+ ions in the lattice. This process creates vacancies in the cation (Na+) sublattice, leading to an increase in the concentration of cation vacancies.
Doping Concentration
- Doping concentration of SrCl2 = 10^-2 mol%
- This means that for every 10000 NaCl units, there are 10 units of SrCl2.
Formation of Cation Vacancies
- Each Sr2+ ion replaces 1 Na+ ion, which creates a cation vacancy.
- Therefore, the number of cation vacancies created will be equal to the number of Sr2+ ions introduced.
Calculating Cation Vacancies
- The concentration of Sr2+ ions can be calculated from the doping concentration.
- 10^-2 mol% = 10^-2/100 × 1 mol = 10^-4 mol of Sr2+ ions per 1 mol of NaCl.
Concentration per Unit Volume
- 1 mole of a substance contains Avogadro's number of entities (approximately 6.023 x 10^23).
- Hence, 10^-4 mol of Sr2+ ions corresponds to:
- 10^-4 mol × 6.023 x 10^23 = 6.023 x 10^19 vacancies per mole of NaCl.
Final Answer
- Therefore, the concentration of cation vacancies in NaCl doped with 10^-2 mol% of SrCl2 is 6.023 x 10^19 mol^-1, corresponding to option 'B'.
This calculation shows that the introduction of SrCl2 significantly affects the structure of NaCl by creating cation vacancies which influence its properties.
When NaCl is doped with SrCl2, the Sr2+ ions replace some Na+ ions in the lattice. This process creates vacancies in the cation (Na+) sublattice, leading to an increase in the concentration of cation vacancies.
Doping Concentration
- Doping concentration of SrCl2 = 10^-2 mol%
- This means that for every 10000 NaCl units, there are 10 units of SrCl2.
Formation of Cation Vacancies
- Each Sr2+ ion replaces 1 Na+ ion, which creates a cation vacancy.
- Therefore, the number of cation vacancies created will be equal to the number of Sr2+ ions introduced.
Calculating Cation Vacancies
- The concentration of Sr2+ ions can be calculated from the doping concentration.
- 10^-2 mol% = 10^-2/100 × 1 mol = 10^-4 mol of Sr2+ ions per 1 mol of NaCl.
Concentration per Unit Volume
- 1 mole of a substance contains Avogadro's number of entities (approximately 6.023 x 10^23).
- Hence, 10^-4 mol of Sr2+ ions corresponds to:
- 10^-4 mol × 6.023 x 10^23 = 6.023 x 10^19 vacancies per mole of NaCl.
Final Answer
- Therefore, the concentration of cation vacancies in NaCl doped with 10^-2 mol% of SrCl2 is 6.023 x 10^19 mol^-1, corresponding to option 'B'.
This calculation shows that the introduction of SrCl2 significantly affects the structure of NaCl by creating cation vacancies which influence its properties.
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If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer?
Question Description
If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer?.
If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer?.
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If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of If NaCl is doped with 10–2mol% of SrCl2, which of the following option shows concentration of cation vacancies?a)10–4mol–1b)6.023 × 1019mol–1c)6.023 × 1021mol–1d)106mol–1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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