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The volume (in mL) of 0.1 M AgNO3 required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'D'. Can you explain this answer?
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Title: Calculation of volume of AgNO3 required for complex precipitation
Given:
- 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2
- 0.1 M AgNO3
To find:
- Volume (in mL) of 0.1 M AgNO3 required for complex precipitation of chloride ions present in the given solution
Solution:
The reaction between [Cr(H2O)5Cl]Cl2 and AgNO3 can be represented as follows:
[Cr(H2O)5Cl]Cl2 + 2AgNO3 → AgCl↓ + [Cr(H2O)5(NO3)]Cl2
The balanced equation shows that 2 moles of AgNO3 react with 1 mole of [Cr(H2O)5Cl]Cl2 to form 1 mole of AgCl precipitate.
Moles of [Cr(H2O)5Cl]Cl2 in 30 mL of 0.01 M solution = 0.01 x 30/1000 = 0.0003 moles
To precipitate all the chloride ions present in the solution, we need 0.0003 moles of AgNO3.
Molarity of AgNO3 = 0.1 M
Moles of AgNO3 required = 0.0003 moles
Volume of AgNO3 required = Moles/Molarity = 0.0003/0.1 = 0.003 L = 3 mL
Therefore, the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is 3 mL or option D.
Given:
- 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2
- 0.1 M AgNO3
To find:
- Volume (in mL) of 0.1 M AgNO3 required for complex precipitation of chloride ions present in the given solution
Solution:
The reaction between [Cr(H2O)5Cl]Cl2 and AgNO3 can be represented as follows:
[Cr(H2O)5Cl]Cl2 + 2AgNO3 → AgCl↓ + [Cr(H2O)5(NO3)]Cl2
The balanced equation shows that 2 moles of AgNO3 react with 1 mole of [Cr(H2O)5Cl]Cl2 to form 1 mole of AgCl precipitate.
Moles of [Cr(H2O)5Cl]Cl2 in 30 mL of 0.01 M solution = 0.01 x 30/1000 = 0.0003 moles
To precipitate all the chloride ions present in the solution, we need 0.0003 moles of AgNO3.
Molarity of AgNO3 = 0.1 M
Moles of AgNO3 required = 0.0003 moles
Volume of AgNO3 required = Moles/Molarity = 0.0003/0.1 = 0.003 L = 3 mL
Therefore, the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is 3 mL or option D.
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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?
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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?.
The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The volume (in mL) of 0.1 M AgNO3required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?.
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