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Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]
Correct answer is between '0.120,0.130'. Can you explain this answer?
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Calculating Vapor Pressure Lowering:
To calculate the vapor pressure lowering caused by the addition of sucrose to water, we can use Raoult's law equation. Raoult's law states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
Formula: ΔP = P°solvent x Xsolute
where ΔP is the decrease in vapor pressure, P°solvent is the vapor pressure of the pure solvent, and Xsolute is the mole fraction of the solute in the solution.
Calculating Mole Fraction:
To calculate the mole fraction of sucrose in the solution, we can use the following formula:
Formula: nsolute/nsolvent + nsolute = 1
where nsolute is the number of moles of sucrose and nsolvent is the number of moles of water.
Calculating Number of Moles:
Number of moles of sucrose = Mass of sucrose/Molecular weight of sucrose
Number of moles of water = Mass of water/Molecular weight of water
Substituting Values:
Number of moles of sucrose = 100 g/342 g/mol = 0.292 moles
Number of moles of water = 1000 g/18 g/mol = 55.56 moles
nsolute/nsolvent + nsolute = 1
0.292/55.56 + 0.292 = 0.0053
Mole fraction of sucrose = 0.0053
Substituting Values:
ΔP = P°solvent x Xsolute
ΔP = 23.8 mmHg x 0.0053
ΔP = 0.126 mmHg
Rounding off:
The answer is between 0.120 and 0.130. Rounding off to 3 decimal places gives us the answer 0.126.
To calculate the vapor pressure lowering caused by the addition of sucrose to water, we can use Raoult's law equation. Raoult's law states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
Formula: ΔP = P°solvent x Xsolute
where ΔP is the decrease in vapor pressure, P°solvent is the vapor pressure of the pure solvent, and Xsolute is the mole fraction of the solute in the solution.
Calculating Mole Fraction:
To calculate the mole fraction of sucrose in the solution, we can use the following formula:
Formula: nsolute/nsolvent + nsolute = 1
where nsolute is the number of moles of sucrose and nsolvent is the number of moles of water.
Calculating Number of Moles:
Number of moles of sucrose = Mass of sucrose/Molecular weight of sucrose
Number of moles of water = Mass of water/Molecular weight of water
Substituting Values:
Number of moles of sucrose = 100 g/342 g/mol = 0.292 moles
Number of moles of water = 1000 g/18 g/mol = 55.56 moles
nsolute/nsolvent + nsolute = 1
0.292/55.56 + 0.292 = 0.0053
Mole fraction of sucrose = 0.0053
Substituting Values:
ΔP = P°solvent x Xsolute
ΔP = 23.8 mmHg x 0.0053
ΔP = 0.126 mmHg
Rounding off:
The answer is between 0.120 and 0.130. Rounding off to 3 decimal places gives us the answer 0.126.
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Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer?
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Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer?.
Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vapor pressure lowering caused by addition of 100 g of Sucrose (M.Wt. = 342g/mol) to 1000 g of water if vapor pressure of pure water at 250C is 23.8 mmHg: [rounded up to three decimal places]Correct answer is between '0.120,0.130'. Can you explain this answer?.
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