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2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .?
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2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . ...
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2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . ...
Law of Multiple Proportions Illustrated:
The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. This law can be illustrated using the given information.
Given Data:
- Mass of metal reacting with AgNO3 to form AgCl: 2.66g
- Mass of AgCl formed: 2.87g
- Mass of another chloride of the same metal reacting with AgNO3: 3.37g
- Mass of AgCl formed in the second reaction: 5.74g
Calculations:
1. Calculate the mass of Cl in the first reaction:
- Mass of AgCl formed = 2.87g
- Mass of Cl = 35.45g/mol (molar mass of Cl)
- Moles of Cl = 2.87 / 35.45 = 0.081 mol
- Mass of Cl in 2.66g of metal = 0.081 * 35.45 = 2.87g
2. Calculate the mass of Cl in the second reaction:
- Mass of AgCl formed = 5.74g
- Moles of Cl = 5.74 / 35.45 = 0.162 mol
- Mass of Cl in 3.37g of metal = 0.162 * 35.45 = 5.74g
Analysis:
- The ratio of Cl to the metal in the two reactions is 1:1. This illustrates the law of multiple proportions because the masses of Cl that combine with a fixed mass of the metal are in a simple ratio of 1:1.
- The two different chlorides of the same metal formed in the reactions have different ratios of Cl to the metal, but these ratios are still simple whole numbers (1:1 and 1:1.5), which is consistent with the law of multiple proportions.
Therefore, the results obtained from the reactions show that they follow the law of multiple proportions, where the ratios of the masses of elements combining with each other are simple whole numbers.
The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. This law can be illustrated using the given information.
Given Data:
- Mass of metal reacting with AgNO3 to form AgCl: 2.66g
- Mass of AgCl formed: 2.87g
- Mass of another chloride of the same metal reacting with AgNO3: 3.37g
- Mass of AgCl formed in the second reaction: 5.74g
Calculations:
1. Calculate the mass of Cl in the first reaction:
- Mass of AgCl formed = 2.87g
- Mass of Cl = 35.45g/mol (molar mass of Cl)
- Moles of Cl = 2.87 / 35.45 = 0.081 mol
- Mass of Cl in 2.66g of metal = 0.081 * 35.45 = 2.87g
2. Calculate the mass of Cl in the second reaction:
- Mass of AgCl formed = 5.74g
- Moles of Cl = 5.74 / 35.45 = 0.162 mol
- Mass of Cl in 3.37g of metal = 0.162 * 35.45 = 5.74g
Analysis:
- The ratio of Cl to the metal in the two reactions is 1:1. This illustrates the law of multiple proportions because the masses of Cl that combine with a fixed mass of the metal are in a simple ratio of 1:1.
- The two different chlorides of the same metal formed in the reactions have different ratios of Cl to the metal, but these ratios are still simple whole numbers (1:1 and 1:1.5), which is consistent with the law of multiple proportions.
Therefore, the results obtained from the reactions show that they follow the law of multiple proportions, where the ratios of the masses of elements combining with each other are simple whole numbers.
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2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .?
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2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .? 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 2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .?.
2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .? 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 2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.66 gram of a metal when treated with AgNO3 gave 2.87 gram of AgCl . 3.37 gram of another chloride of the same metal gave 5.74 gram of AgCl when treated with AgNO3 solution . Show that the results illustrates law of multiple proportions .?.
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