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Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg?
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Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The ...
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Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The ...
Density of the Solution
Given:
Density of the 2.05 M solution of acetic acid in water = 1.02 g/mL
The density of a solution is defined as the ratio of the mass of the solution to its volume. In this case, we are given the density of the solution, which is 1.02 g/mL.
Calculating the Mass of the Solution
To calculate the mass of the solution, we need to know its volume. Since the density is given in g/mL, we can assume a volume of 1 mL for simplicity.
Therefore, the mass of the solution = density x volume
= 1.02 g/mL x 1 mL
= 1.02 g
Calculating the Mass of Acetic Acid
The concentration of the solution is given as 2.05 M, which means there are 2.05 moles of acetic acid in 1 liter of the solution.
To calculate the mass of acetic acid in the solution, we need to know the molar mass of acetic acid. The molar mass of acetic acid (CH3COOH) is 60.052 g/mol.
Therefore, the mass of acetic acid = molar mass x moles
= 60.052 g/mol x 2.05 mol
= 123.061 g
Calculating the Mass of Water
To find the mass of water in the solution, we subtract the mass of acetic acid from the mass of the solution.
Mass of water = mass of solution - mass of acetic acid
= 1.02 g - 123.061 g
= -122.041 g
Since the mass of water is negative, it indicates that there is no water present in the solution. This suggests that the given density of the solution is incorrect.
Therefore, we cannot determine the molality of the solution based on the given information.
Conclusion
Based on the given density of the solution, it is not possible to calculate the molality of the acetic acid solution in water. The negative mass of water indicates that the given density is incorrect. More information or a correct density value is required to determine the molality of the solution accurately.
Given:
Density of the 2.05 M solution of acetic acid in water = 1.02 g/mL
The density of a solution is defined as the ratio of the mass of the solution to its volume. In this case, we are given the density of the solution, which is 1.02 g/mL.
Calculating the Mass of the Solution
To calculate the mass of the solution, we need to know its volume. Since the density is given in g/mL, we can assume a volume of 1 mL for simplicity.
Therefore, the mass of the solution = density x volume
= 1.02 g/mL x 1 mL
= 1.02 g
Calculating the Mass of Acetic Acid
The concentration of the solution is given as 2.05 M, which means there are 2.05 moles of acetic acid in 1 liter of the solution.
To calculate the mass of acetic acid in the solution, we need to know the molar mass of acetic acid. The molar mass of acetic acid (CH3COOH) is 60.052 g/mol.
Therefore, the mass of acetic acid = molar mass x moles
= 60.052 g/mol x 2.05 mol
= 123.061 g
Calculating the Mass of Water
To find the mass of water in the solution, we subtract the mass of acetic acid from the mass of the solution.
Mass of water = mass of solution - mass of acetic acid
= 1.02 g - 123.061 g
= -122.041 g
Since the mass of water is negative, it indicates that there is no water present in the solution. This suggests that the given density of the solution is incorrect.
Therefore, we cannot determine the molality of the solution based on the given information.
Conclusion
Based on the given density of the solution, it is not possible to calculate the molality of the acetic acid solution in water. The negative mass of water indicates that the given density is incorrect. More information or a correct density value is required to determine the molality of the solution accurately.
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Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg?
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Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg?.
Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Density of 2.0 5 M Solution of acetic acid in water is 1.02 g/mL. The molality of the solution is 1)0.44 mol/kg 2)1.14 mol/kg 3)3.28 mol/kg 4)2.29 mol/kg?.
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