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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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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of...
[Cr(H20)5Cl]Cl2+2AgNO3 =》2AgCl + [Cr(H20)5Cl](NO3)2
so 2 ionizable cl- ion present . further steps will be just equating 2×0.01×30=0.1×V and Vcomes out to be 6
so 2 ionizable cl- ion present . further steps will be just equating 2×0.01×30=0.1×V and Vcomes out to be 6
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The volume (in mL) of 0.1 M AgNO3required for complex precipitation of...
Problem:
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
Solution:
To determine the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions, we need to calculate the amount of chloride ions present in the 30 mL of 0.01 M [Cr(H2O)5Cl]Cl2 solution and then use the stoichiometry of the reaction to determine the volume of AgNO3 required.
Step 1: Calculate the amount of chloride ions in the [Cr(H2O)5Cl]Cl2 solution
Given:
Volume of [Cr(H2O)5Cl]Cl2 solution = 30 mL
Concentration of [Cr(H2O)5Cl]Cl2 solution = 0.01 M
The amount of chloride ions present in the solution can be calculated using the formula:
Amount of chloride ions = Concentration of solution x Volume of solution
Amount of chloride ions = 0.01 M x 30 mL
Amount of chloride ions = 0.01 mol/L x 0.03 L
Amount of chloride ions = 0.0003 mol
Step 2: Use stoichiometry to determine the volume of AgNO3
The balanced chemical equation for the reaction between silver nitrate (AgNO3) and chloride ions (Cl-) is:
AgNO3 + Cl- -> AgCl + NO3-
From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl-. Therefore, the amount of AgNO3 required to react with the chloride ions can be calculated as:
Amount of AgNO3 = Amount of chloride ions
Amount of AgNO3 = 0.0003 mol
Now, we can use the formula for concentration to calculate the volume of AgNO3:
Volume of AgNO3 = Amount of AgNO3 / Concentration of AgNO3
Volume of AgNO3 = 0.0003 mol / 0.1 mol/L
Volume of AgNO3 = 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 [Cr(H2O)5Cl]Cl2 solution is close to 3 mL.
Hence, the correct answer is option 'a) 3'.
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
Solution:
To determine the volume of 0.1 M AgNO3 required for complex precipitation of chloride ions, we need to calculate the amount of chloride ions present in the 30 mL of 0.01 M [Cr(H2O)5Cl]Cl2 solution and then use the stoichiometry of the reaction to determine the volume of AgNO3 required.
Step 1: Calculate the amount of chloride ions in the [Cr(H2O)5Cl]Cl2 solution
Given:
Volume of [Cr(H2O)5Cl]Cl2 solution = 30 mL
Concentration of [Cr(H2O)5Cl]Cl2 solution = 0.01 M
The amount of chloride ions present in the solution can be calculated using the formula:
Amount of chloride ions = Concentration of solution x Volume of solution
Amount of chloride ions = 0.01 M x 30 mL
Amount of chloride ions = 0.01 mol/L x 0.03 L
Amount of chloride ions = 0.0003 mol
Step 2: Use stoichiometry to determine the volume of AgNO3
The balanced chemical equation for the reaction between silver nitrate (AgNO3) and chloride ions (Cl-) is:
AgNO3 + Cl- -> AgCl + NO3-
From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl-. Therefore, the amount of AgNO3 required to react with the chloride ions can be calculated as:
Amount of AgNO3 = Amount of chloride ions
Amount of AgNO3 = 0.0003 mol
Now, we can use the formula for concentration to calculate the volume of AgNO3:
Volume of AgNO3 = Amount of AgNO3 / Concentration of AgNO3
Volume of AgNO3 = 0.0003 mol / 0.1 mol/L
Volume of AgNO3 = 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 [Cr(H2O)5Cl]Cl2 solution is close to 3 mL.
Hence, the correct answer is option 'a) 3'.
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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?
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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 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 toa)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 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 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 toa)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 toa)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?.
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