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1.84 g of a dibromide of a metal M on reaction with excess of AgNO, 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 AgNO, ga...
Given Data:
- Mass of dibromide of metal M = 1.84 g
- Mass of yellow precipitate (AgBr) formed = 3.76 g
- Molar mass of Br = 80 g/mol
- Molar mass of Ag = 108 g/mol
Calculations:
Step 1: Calculate the moles of AgBr formed
- The molar mass of AgBr = Ag (108 g/mol) + Br (80 g/mol) = 188 g/mol
- Moles of AgBr formed = Mass of AgBr / Molar mass of AgBr
= 3.76 g / 188 g/mol
= 0.02 mol
Step 2: Determine the moles of Br in AgBr
- Since AgBr contains 1 Ag and 1 Br, moles of Br = moles of AgBr
- Moles of Br = 0.02 mol
Step 3: Calculate the moles of M in the dibromide
- Molar ratio of Br to M in the dibromide = 2:1 (from the formula MBr2)
- Moles of M = 0.02 mol / 2
= 0.01 mol
Step 4: Determine the molar mass of M
- Mass of M = Moles of M x Molar mass of M
= 0.01 mol x m
= m
Therefore, the molar mass of the metal M in the dibromide is equal to the mass of the metal itself, which is 'm'.
- Mass of dibromide of metal M = 1.84 g
- Mass of yellow precipitate (AgBr) formed = 3.76 g
- Molar mass of Br = 80 g/mol
- Molar mass of Ag = 108 g/mol
Calculations:
Step 1: Calculate the moles of AgBr formed
- The molar mass of AgBr = Ag (108 g/mol) + Br (80 g/mol) = 188 g/mol
- Moles of AgBr formed = Mass of AgBr / Molar mass of AgBr
= 3.76 g / 188 g/mol
= 0.02 mol
Step 2: Determine the moles of Br in AgBr
- Since AgBr contains 1 Ag and 1 Br, moles of Br = moles of AgBr
- Moles of Br = 0.02 mol
Step 3: Calculate the moles of M in the dibromide
- Molar ratio of Br to M in the dibromide = 2:1 (from the formula MBr2)
- Moles of M = 0.02 mol / 2
= 0.01 mol
Step 4: Determine the molar mass of M
- Mass of M = Moles of M x Molar mass of M
= 0.01 mol x m
= m
Therefore, the molar mass of the metal M in the dibromide is equal to the mass of the metal itself, which is 'm'.
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1.84 g of a dibromide of a metal M on reaction with excess of AgNO, 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 AgNO, 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 AgNO, 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 AgNO, 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 AgNO, 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 AgNO, 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 AgNO, gave 3.76 g of yellow precipitate. Thus, m of metal dibromide is (Br80, Ag=108)?.
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