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Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010]
- a)5.0 x 10-8 g
- b)1.2 x 10-10 g
- c)1.2 x 10-9 g
- d)6.2 x 10-5 g
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Solubility product of silver bromide is 5.0 x 10-13. The quantity of p...
For precipitation
[Ag+][Br-] > Ksp[AgBr]
[Br-]min = (5 x 10-13)/0.05 = 10-11 M
Mass of potassium bromide needed = 10-11 x 120
= 1.2 x 10-9 g
Most Upvoted Answer
Solubility product of silver bromide is 5.0 x 10-13. The quantity of p...
To start the precipitation of AgBr, we need to add an appropriate amount of potassium bromide (KBr) to a solution of silver nitrate (AgNO3). The solubility product of silver bromide (AgBr) is given as 5.0 x 10^-13.
Let's calculate the number of moles of Ag+ ions in 1 liter of 0.05 M silver nitrate solution.
Number of moles of Ag+ = Molarity x Volume
= 0.05 mol/L x 1 L
= 0.05 mol
Since the solubility product of AgBr is given as 5.0 x 10^-13, we can write the equilibrium expression for the precipitation of AgBr as follows:
Ag+ + Br- → AgBr
According to the stoichiometry of the reaction, 1 mole of Ag+ reacts with 1 mole of Br-. Therefore, the concentration of Br- ions in the solution will be equal to the concentration of Ag+ ions, which is 0.05 mol/L.
Now, let's calculate the number of moles of KBr required to provide the necessary Br- ions.
Number of moles of KBr = Moles of Br- ions required
= 0.05 mol
The molar mass of KBr is given as 120 g/mol. Using this information, we can calculate the mass of KBr required.
Mass of KBr = Number of moles x Molar mass
= 0.05 mol x 120 g/mol
= 6 g
However, we need to convert this mass to grams because the options are given in grams. The conversion factor between grams and micrograms is 1 g = 10^6 µg.
Therefore, the mass of KBr required in micrograms is:
Mass of KBr = 6 g x 10^6 µg/g
= 6 x 10^6 µg
Comparing this with the given options, we can see that option C is the closest value:
Option C) 1.2 x 10^-9 g
Hence, the correct answer is option C) 1.2 x 10^-9 g.
Let's calculate the number of moles of Ag+ ions in 1 liter of 0.05 M silver nitrate solution.
Number of moles of Ag+ = Molarity x Volume
= 0.05 mol/L x 1 L
= 0.05 mol
Since the solubility product of AgBr is given as 5.0 x 10^-13, we can write the equilibrium expression for the precipitation of AgBr as follows:
Ag+ + Br- → AgBr
According to the stoichiometry of the reaction, 1 mole of Ag+ reacts with 1 mole of Br-. Therefore, the concentration of Br- ions in the solution will be equal to the concentration of Ag+ ions, which is 0.05 mol/L.
Now, let's calculate the number of moles of KBr required to provide the necessary Br- ions.
Number of moles of KBr = Moles of Br- ions required
= 0.05 mol
The molar mass of KBr is given as 120 g/mol. Using this information, we can calculate the mass of KBr required.
Mass of KBr = Number of moles x Molar mass
= 0.05 mol x 120 g/mol
= 6 g
However, we need to convert this mass to grams because the options are given in grams. The conversion factor between grams and micrograms is 1 g = 10^6 µg.
Therefore, the mass of KBr required in micrograms is:
Mass of KBr = 6 g x 10^6 µg/g
= 6 x 10^6 µg
Comparing this with the given options, we can see that option C is the closest value:
Option C) 1.2 x 10^-9 g
Hence, the correct answer is option C) 1.2 x 10^-9 g.
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Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer?
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Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer?.
Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is [AIEEE-2010] a)5.0 x 10-8 gb)1.2 x 10-10 gc)1.2 x 10-9 gd)6.2 x 10-5 gCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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