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When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is?
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When a perfect monolayer of steric acid is formed in the air water int...
Length of the molecule of stearic acid at the air-water interface:
Given data:
- Molecular weight of stearic acid (M) = 284 g/mol
- Density of stearic acid (ρ) = 0.94 g/cm³
- Area occupied by each stearic acid molecule (A) = 20 Ų
To find the length of the stearic acid molecule, we can use the following formula:
Volume (V) = Mass (m) / Density (ρ)
Since the area occupied by each stearic acid molecule is given, we can calculate the volume occupied as follows:
Volume (V) = Area (A) * Height (h)
where h is the height or length of the molecule.
We can rearrange the formula to solve for the length of the molecule:
h = V / A
To calculate the volume (V), we need to find the mass (m) of the stearic acid molecule. The mass can be calculated using the molecular weight (M) of stearic acid:
m = M / Avogadro's number
Calculations:
1. Calculate the mass of one stearic acid molecule:
- m = 284 g/mol / 6.022 x 10²³ molecules/mol (Avogadro's number)
- m ≈ 4.72 x 10⁻²³ g
2. Calculate the volume occupied by one stearic acid molecule:
- V = m / ρ
- V = 4.72 x 10⁻²³ g / 0.94 g/cm³
- V ≈ 5.02 x 10⁻²³ cm³
3. Convert the volume to cubic angstroms:
- 1 cm³ = 10⁈⁸ ų
- V ≈ 5.02 x 10⁻²³ cm³ * 10⁈⁸ ų/cm³
- V ≈ 5.02 x 10⁻¹⁵ ų
4. Calculate the length of the molecule:
- h = V / A
- h = 5.02 x 10⁻¹⁵ ų / 20 Ų
- h ≈ 2.51 x 10⁻¹⁵ Å
Conclusion:
The length of the stearic acid molecule at the air-water interface is approximately 2.51 x 10⁻¹⁵ Å.
Given data:
- Molecular weight of stearic acid (M) = 284 g/mol
- Density of stearic acid (ρ) = 0.94 g/cm³
- Area occupied by each stearic acid molecule (A) = 20 Ų
To find the length of the stearic acid molecule, we can use the following formula:
Volume (V) = Mass (m) / Density (ρ)
Since the area occupied by each stearic acid molecule is given, we can calculate the volume occupied as follows:
Volume (V) = Area (A) * Height (h)
where h is the height or length of the molecule.
We can rearrange the formula to solve for the length of the molecule:
h = V / A
To calculate the volume (V), we need to find the mass (m) of the stearic acid molecule. The mass can be calculated using the molecular weight (M) of stearic acid:
m = M / Avogadro's number
Calculations:
1. Calculate the mass of one stearic acid molecule:
- m = 284 g/mol / 6.022 x 10²³ molecules/mol (Avogadro's number)
- m ≈ 4.72 x 10⁻²³ g
2. Calculate the volume occupied by one stearic acid molecule:
- V = m / ρ
- V = 4.72 x 10⁻²³ g / 0.94 g/cm³
- V ≈ 5.02 x 10⁻²³ cm³
3. Convert the volume to cubic angstroms:
- 1 cm³ = 10⁈⁸ ų
- V ≈ 5.02 x 10⁻²³ cm³ * 10⁈⁸ ų/cm³
- V ≈ 5.02 x 10⁻¹⁵ ų
4. Calculate the length of the molecule:
- h = V / A
- h = 5.02 x 10⁻¹⁵ ų / 20 Ų
- h ≈ 2.51 x 10⁻¹⁵ Å
Conclusion:
The length of the stearic acid molecule at the air-water interface is approximately 2.51 x 10⁻¹⁵ Å.
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When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is?
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When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is?.
When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a perfect monolayer of steric acid is formed in the air water interface each molecular steric acid molecular weight 284 density 0.94 gram per centimetre cube occupies an area of 20 angstrom square the length in angstrom of the molecule is?.
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