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1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)?
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1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 ga...
Given Data:
- Mass of dibromide of metal M = 1.84 g
- Mass of yellow precipitate formed (AgBr) = 3.76 g
- Atomic masses: Br = 80, Ag = 108
Calculation:
1. Determine the mass of bromine in the precipitate:
- The molar mass of AgBr = Ag (108) + Br (80) = 188 g/mol
- Number of moles of AgBr formed = Mass of AgBr / Molar mass of AgBr
= 3.76 g / 188 g/mol = 0.02 mol
- Since 1 mole of AgBr contains 1 mole of bromine, the mass of bromine = 0.02 mol x 80 g/mol = 1.6 g
2. Calculate the mass of metal M in the dibromide:
- Mass of metal M = Mass of dibromide - Mass of bromine
= 1.84 g - 1.6 g = 0.24 g
Conclusion:
- The mass of metal M in the dibromide compound is 0.24 g.
- Mass of dibromide of metal M = 1.84 g
- Mass of yellow precipitate formed (AgBr) = 3.76 g
- Atomic masses: Br = 80, Ag = 108
Calculation:
1. Determine the mass of bromine in the precipitate:
- The molar mass of AgBr = Ag (108) + Br (80) = 188 g/mol
- Number of moles of AgBr formed = Mass of AgBr / Molar mass of AgBr
= 3.76 g / 188 g/mol = 0.02 mol
- Since 1 mole of AgBr contains 1 mole of bromine, the mass of bromine = 0.02 mol x 80 g/mol = 1.6 g
2. Calculate the mass of metal M in the dibromide:
- Mass of metal M = Mass of dibromide - Mass of bromine
= 1.84 g - 1.6 g = 0.24 g
Conclusion:
- The mass of metal M in the dibromide compound is 0.24 g.
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1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)?
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1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)?.
1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.84 g of a dibromide of a metal M on reaction with excess of AgNO3 gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)?.
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