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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...
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) X 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...
Given:
- Volume of [Cr(H2O)5Cl]Cl2 solution = 30 mL
- Concentration [Cr(H2O)5Cl]Cl2 solution = 0.01 M
- Concentration of AgNO3 solution = 0.1 M
To precipitate chloride ions present in the [Cr(H2O)5Cl]Cl2 solution, we need to add AgNO3 solution, which will react with chloride ions to form silver chloride precipitate.
The balanced chemical equation for the reaction is:
AgNO3 + NaCl → AgCl + NaNO3
From the equation, we can see that 1 mole of AgNO3 will react with 1 mole of NaCl (or chloride ions) to form 1 mole of AgCl.
To determine the volume of AgNO3 solution required for precipitation, we need to calculate the number of moles of chloride ions in the [Cr(H2O)5Cl]Cl2 solution:
Number of moles of [Cr(H2O)5Cl]Cl2 = concentration x volume = 0.01 x 0.03 = 0.0003 moles
Number of moles of Cl- ions = 0.0003 x 2 = 0.0006 moles (since there are 2 chloride ions per molecule of [Cr(H2O)5Cl]Cl2)
Now, we can calculate the number of moles of AgNO3 required to react with the chloride ions:
Number of moles of AgNO3 = number of moles of Cl- ions = 0.0006 moles
Finally, we can calculate the volume of AgNO3 solution required using the concentration:
Volume of AgNO3 = number of moles / concentration = 0.0006 / 0.1 = 0.006 L = 6 mL
Therefore, the correct option is (d) 6 mL.
- Volume of [Cr(H2O)5Cl]Cl2 solution = 30 mL
- Concentration [Cr(H2O)5Cl]Cl2 solution = 0.01 M
- Concentration of AgNO3 solution = 0.1 M
To precipitate chloride ions present in the [Cr(H2O)5Cl]Cl2 solution, we need to add AgNO3 solution, which will react with chloride ions to form silver chloride precipitate.
The balanced chemical equation for the reaction is:
AgNO3 + NaCl → AgCl + NaNO3
From the equation, we can see that 1 mole of AgNO3 will react with 1 mole of NaCl (or chloride ions) to form 1 mole of AgCl.
To determine the volume of AgNO3 solution required for precipitation, we need to calculate the number of moles of chloride ions in the [Cr(H2O)5Cl]Cl2 solution:
Number of moles of [Cr(H2O)5Cl]Cl2 = concentration x volume = 0.01 x 0.03 = 0.0003 moles
Number of moles of Cl- ions = 0.0003 x 2 = 0.0006 moles (since there are 2 chloride ions per molecule of [Cr(H2O)5Cl]Cl2)
Now, we can calculate the number of moles of AgNO3 required to react with the chloride ions:
Number of moles of AgNO3 = number of moles of Cl- ions = 0.0006 moles
Finally, we can calculate the volume of AgNO3 solution required using the concentration:
Volume of AgNO3 = number of moles / concentration = 0.0006 / 0.1 = 0.006 L = 6 mL
Therefore, the correct option is (d) 6 mL.
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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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