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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?
Verified Answer
The volume (in mL) of 0.1 M AgNO3required for complex precipitation of...
The chemical reaction between them is...[Cr(H2O)5Cl]Cl2 + {2}AgNO3 ——> {2}AgCl(↓) + [Cr(H2O)5Cl]^2+ + {2}NO3-moles of coordinate compound given = molarity x volume(in litre)mole = 0.01 x (30/1000) = (3/10000)for one mole coordinate compound.. we need 2 mole AgNO3....i.e.,1mole complex ===> 2 mole AgNO3(3/10000) mole ==> (2x3/10000) mole AgNO3molarity = moles/volume (in litre)volume(in litre) = mole / molarityV= (6/10000)/0.1 = (0.6) Litrehence volume required = 0.6litre = 6ml
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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of...
Understanding the Problem
To determine the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions from the given solution, we need to analyze the components involved.
Components of the Solution
- [Cr(H2O)5Cl]Cl2: This complex contains two chloride ions (Cl-) for each formula unit.
- Volume of the solution: 30 mL (which is 0.030 L).
- Concentration of the complex: 0.01 M.
Calculating moles of Chloride Ions
- Moles of [Cr(H2O)5Cl]Cl2 in 30 mL:
- Moles = Concentration × Volume
- Moles = 0.01 M × 0.030 L = 0.0003 moles.
- Since there are 2 chloride ions per complex, total moles of Cl-:
- Total moles = 0.0003 moles × 2 = 0.0006 moles of Cl-.
Determining Volume of AgNO3 Required
- The reaction for precipitation is:
- AgNO3 + Cl- → AgCl (s).
- This means that 1 mole of AgNO3 is required for 1 mole of Cl-. Hence, we need 0.0006 moles of AgNO3.
- Volume of 0.1 M AgNO3 needed:
- Volume = Moles / Concentration
- Volume = 0.0006 moles / 0.1 M = 0.006 L = 6 mL.
Conclusion
The volume of 0.1 M AgNO3 required for the complete precipitation of the chloride ions in the given solution is approximately 6 mL. Therefore, the correct answer is option 'D'.
To determine the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions from the given solution, we need to analyze the components involved.
Components of the Solution
- [Cr(H2O)5Cl]Cl2: This complex contains two chloride ions (Cl-) for each formula unit.
- Volume of the solution: 30 mL (which is 0.030 L).
- Concentration of the complex: 0.01 M.
Calculating moles of Chloride Ions
- Moles of [Cr(H2O)5Cl]Cl2 in 30 mL:
- Moles = Concentration × Volume
- Moles = 0.01 M × 0.030 L = 0.0003 moles.
- Since there are 2 chloride ions per complex, total moles of Cl-:
- Total moles = 0.0003 moles × 2 = 0.0006 moles of Cl-.
Determining Volume of AgNO3 Required
- The reaction for precipitation is:
- AgNO3 + Cl- → AgCl (s).
- This means that 1 mole of AgNO3 is required for 1 mole of Cl-. Hence, we need 0.0006 moles of AgNO3.
- Volume of 0.1 M AgNO3 needed:
- Volume = Moles / Concentration
- Volume = 0.0006 moles / 0.1 M = 0.006 L = 6 mL.
Conclusion
The volume of 0.1 M AgNO3 required for the complete precipitation of the chloride ions in the given solution is approximately 6 mL. Therefore, the correct answer is option 'D'.
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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of...
N1V1=N2V2. 0.1*V1 2*0.01*30. V1 = 6
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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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 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 toa)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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 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 toa)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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?, a detailed solution 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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of 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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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