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Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.
Q.
The % of unprecipitated silver ions is
- a)0.15%
- b)0.37%
- c)0.67%
- d)0.97%
Correct answer is option 'C'. Can you explain this answer?
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Radiochemical techniques can be used to determine solubility product e...
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To determine the % of unprecipitated silver ions, we need to understand the process described in the given experiment and then calculate the concentration of silver ions in the final solution.
1. Initial Conditions:
- Initial volume of Ag+ solution (V1) = 50.00 ml
- Initial concentration of Ag+ solution (C1) = 0.010 81 A * 10^3 M
- Initial radioactivity of Ag+ solution = 75,000 counts per min per ml
2. Mixing the Solutions:
- 100 ml of 0.03 M NaIO3 solution is mixed with the Ag+ solution.
- This results in the formation of AgIO3 precipitate according to the reaction: Ag+ + IO3- → AgIO3
3. Dilution and Filtration:
- The resulting mixed solution is diluted to a final volume of 500 ml.
- The solution is then filtered to remove all the AgIO3 precipitate, leaving behind a radioactive solution.
4. Calculation of Concentration:
- We need to calculate the concentration of Ag+ ions in the final solution after the precipitation.
- Since AgIO3 is a sparingly soluble salt, we assume that the solubility product (Ksp) is not exceeded.
- Let x be the concentration of Ag+ ions in the final solution.
5. Applying the Law of Mass Action:
- The solubility product expression for AgIO3 is given by: Ksp = [Ag+][IO3-]
- Since IO3- is in excess, we can assume its concentration remains constant.
- Therefore, [Ag+] = Ksp / [IO3-] = Ksp / (0.03 M) = Ksp / 30
6. Determining Ksp:
- The molar mass of AgIO3 is given as 285 g/mol.
- Since the solution is diluted to a final volume of 500 ml, the moles of AgIO3 present in the solution can be calculated as: moles = (mass / molar mass) * (1 / volume) = (285 * x) / 500
- The moles of AgIO3 can also be calculated from the initial Ag+ solution: moles = C1 * (V1 / 1000)
- Equating these two expressions for moles, we get: (285 * x) / 500 = C1 * (V1 / 1000)
7. Substituting Values and Solving:
- Substituting the given values, we have: (285 * x) / 500 = (0.01081 * 50) / 1000
- Solving for x, we find: x = (0.01081 * 50 * 500) / (1000 * 285) ≈ 0.0095 M
8. Calculating the % of Unprecipitated Silver Ions:
- The % of unprecipitated silver ions can be calculated as: (% unprecipitated Ag+) = (x / C1) * 100 = (0.0095 / 0.01081) * 100 ≈ 87.7%
- However, we need to subtract this value from 100% to find the % of precipitated silver ions.
- Therefore, % of unprecipitated silver ions = 100% -
1. Initial Conditions:
- Initial volume of Ag+ solution (V1) = 50.00 ml
- Initial concentration of Ag+ solution (C1) = 0.010 81 A * 10^3 M
- Initial radioactivity of Ag+ solution = 75,000 counts per min per ml
2. Mixing the Solutions:
- 100 ml of 0.03 M NaIO3 solution is mixed with the Ag+ solution.
- This results in the formation of AgIO3 precipitate according to the reaction: Ag+ + IO3- → AgIO3
3. Dilution and Filtration:
- The resulting mixed solution is diluted to a final volume of 500 ml.
- The solution is then filtered to remove all the AgIO3 precipitate, leaving behind a radioactive solution.
4. Calculation of Concentration:
- We need to calculate the concentration of Ag+ ions in the final solution after the precipitation.
- Since AgIO3 is a sparingly soluble salt, we assume that the solubility product (Ksp) is not exceeded.
- Let x be the concentration of Ag+ ions in the final solution.
5. Applying the Law of Mass Action:
- The solubility product expression for AgIO3 is given by: Ksp = [Ag+][IO3-]
- Since IO3- is in excess, we can assume its concentration remains constant.
- Therefore, [Ag+] = Ksp / [IO3-] = Ksp / (0.03 M) = Ksp / 30
6. Determining Ksp:
- The molar mass of AgIO3 is given as 285 g/mol.
- Since the solution is diluted to a final volume of 500 ml, the moles of AgIO3 present in the solution can be calculated as: moles = (mass / molar mass) * (1 / volume) = (285 * x) / 500
- The moles of AgIO3 can also be calculated from the initial Ag+ solution: moles = C1 * (V1 / 1000)
- Equating these two expressions for moles, we get: (285 * x) / 500 = C1 * (V1 / 1000)
7. Substituting Values and Solving:
- Substituting the given values, we have: (285 * x) / 500 = (0.01081 * 50) / 1000
- Solving for x, we find: x = (0.01081 * 50 * 500) / (1000 * 285) ≈ 0.0095 M
8. Calculating the % of Unprecipitated Silver Ions:
- The % of unprecipitated silver ions can be calculated as: (% unprecipitated Ag+) = (x / C1) * 100 = (0.0095 / 0.01081) * 100 ≈ 87.7%
- However, we need to subtract this value from 100% to find the % of precipitated silver ions.
- Therefore, % of unprecipitated silver ions = 100% -
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Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer?
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Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. 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 Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer?.
Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. 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 Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer?.
Solutions for Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Radiochemical techniques can be used to determine solubility product eslimation. The measurement of radioactivity can be used to find the concentration in a solubthty equilibrium giving afak idea about various equilibrium concentrations. In an experiment, 50.00m1 of a 0.010 81A*1103sdufion containing a silver isotope with a radioactivity of 75,000 counts per min per mi were mixed viii 100 ml of a 0.03 Na103 solution. The mixed solution was diluted 500 mi and filtered to remove al oftheAg/03predpilateleavinig behind a radioactive solution. Molar mass of AgI03 = 285 g/mol.Q.The % of unprecipitated silver ions isa)0.15%b)0.37%c)0.67%d)0.97%Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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