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In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. The percentage of bromine in the compound is:
- a)43%
- b)34.04%
- c)34%
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In Carius method of estimation of halogen, 0.15 g of an organic compou...
Molar mass of AgBr = 108 + 80 = 188 g mol-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains= 80 X 0.12/188 g bromine
Percentage of bromine=80 X0.12 X100/188 X 0.15
= 34.04%
Most Upvoted Answer
In Carius method of estimation of halogen, 0.15 g of an organic compou...
Explanation:
Carius Method: In this method, an organic compound is heated with fuming nitric acid in a hard glass tube called a Carius tube. The halogen present in the organic compound is converted into halogen acid which is then precipitated as silver halide on the addition of silver nitrate solution. The weight of the silver halide formed is then used to calculate the percentage of halogen in the organic compound.
Given, the weight of organic compound = 0.15 g
Weight of AgBr formed = 0.12 g
Let us calculate the percentage of bromine in the compound using the formula:
Percentage of bromine = (Weight of AgBr formed / Weight of organic compound) x 100
Substituting the given values, we get
Percentage of bromine = (0.12 / 0.15) x 100 = 80%
But we know that the molecular weight of AgBr is 187 g/mol, i.e., 107 g/mol for Ag and 80 g/mol for Br.
Therefore, the number of moles of AgBr formed = 0.12 / 187 = 0.0006417 mol
The number of moles of bromine in the compound = 0.0006417 x 1 = 0.0006417 mol
The molecular weight of the compound can be calculated using the formula:
Molecular weight = (Weight of organic compound / Number of moles of organic compound)
Substituting the given values, we get
Molecular weight = (0.15 / 0.0006417) = 233.423 g/mol
The molecular formula of the compound can be calculated using the empirical formula and the molecular weight. Since the compound contains bromine, we can assume the empirical formula to be CxHyBrz.
The empirical formula weight = 12x + 1y + 80z
The molecular weight = n(12x + 1y + 80z), where n is an integer.
Substituting the values, we get
233.423 = n(12x + y + 80z)
Since we do not know the values of x, y, and z, we cannot solve this equation. However, we can assume a value of x and calculate the values of y and z.
Assuming x = 10, we get
12x + y + 80z = 233.423 / n = 2334.23 / 10n
Solving for y and z, we get
y = 2334.23 / 10n - 120
z = (0.12 / 187) / n = 0.0006417 / n
Since y and z should be integers, we can assume n = 10 and
Carius Method: In this method, an organic compound is heated with fuming nitric acid in a hard glass tube called a Carius tube. The halogen present in the organic compound is converted into halogen acid which is then precipitated as silver halide on the addition of silver nitrate solution. The weight of the silver halide formed is then used to calculate the percentage of halogen in the organic compound.
Given, the weight of organic compound = 0.15 g
Weight of AgBr formed = 0.12 g
Let us calculate the percentage of bromine in the compound using the formula:
Percentage of bromine = (Weight of AgBr formed / Weight of organic compound) x 100
Substituting the given values, we get
Percentage of bromine = (0.12 / 0.15) x 100 = 80%
But we know that the molecular weight of AgBr is 187 g/mol, i.e., 107 g/mol for Ag and 80 g/mol for Br.
Therefore, the number of moles of AgBr formed = 0.12 / 187 = 0.0006417 mol
The number of moles of bromine in the compound = 0.0006417 x 1 = 0.0006417 mol
The molecular weight of the compound can be calculated using the formula:
Molecular weight = (Weight of organic compound / Number of moles of organic compound)
Substituting the given values, we get
Molecular weight = (0.15 / 0.0006417) = 233.423 g/mol
The molecular formula of the compound can be calculated using the empirical formula and the molecular weight. Since the compound contains bromine, we can assume the empirical formula to be CxHyBrz.
The empirical formula weight = 12x + 1y + 80z
The molecular weight = n(12x + 1y + 80z), where n is an integer.
Substituting the values, we get
233.423 = n(12x + y + 80z)
Since we do not know the values of x, y, and z, we cannot solve this equation. However, we can assume a value of x and calculate the values of y and z.
Assuming x = 10, we get
12x + y + 80z = 233.423 / n = 2334.23 / 10n
Solving for y and z, we get
y = 2334.23 / 10n - 120
z = (0.12 / 187) / n = 0.0006417 / n
Since y and z should be integers, we can assume n = 10 and
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In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. The percentage of bromine in the compound is:a)43%b)34.04%c)34%d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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