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what weight of glucose should be added to 1700g of water at 20 degree celsius to lower its vapour pressure by 0.01mm the vapour pressure of water at 20 degree celsius 17mmhg
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what weight of glucose should be added to 1700g of water at 20 degree ...
Understanding the Problem
To determine the weight of glucose required to lower the vapor pressure of water by 0.01 mmHg, we can use Raoult's Law. This states that the vapor pressure of a solution is directly related to the mole fraction of the solvent.
Given Data
- Vapor pressure of pure water at 20°C: 17 mmHg
- Desired reduction in vapor pressure: 0.01 mmHg
- Mass of water: 1700 g
Calculating the Reduction in Vapor Pressure
- Final vapor pressure of the solution = 17 mmHg - 0.01 mmHg = 16.99 mmHg
Calculating Mole Fraction of Water
Using Raoult's Law:
- ΔP = P° * X_solute
Where:
- ΔP = change in vapor pressure
- P° = vapor pressure of pure solvent (17 mmHg)
- X_solute = mole fraction of solute (glucose)
Rearranging the formula gives us:
- X_solute = ΔP / P°
Substituting values:
- X_solute = 0.01 mmHg / 17 mmHg ≈ 0.000588
Calculating Moles of Water
- Molar mass of water = 18 g/mol
- Moles of water = 1700 g / 18 g/mol ≈ 94.44 moles
Finding Moles of Glucose
Since X_solute = moles of glucose / (moles of glucose + moles of water):
- Let moles of glucose = y
- Thus, X_solute = y / (y + 94.44)
Setting up the equation:
- 0.000588 = y / (y + 94.44)
Solving for y gives approximately:
- y ≈ 0.055 moles of glucose
Calculating Weight of Glucose
- Molar mass of glucose (C6H12O6) ≈ 180 g/mol
- Weight of glucose = 0.055 moles * 180 g/mol ≈ 9.9 g
Conclusion
To lower the vapor pressure of 1700 g of water by 0.01 mmHg at 20°C, approximately 9.9 grams of glucose should be added.
To determine the weight of glucose required to lower the vapor pressure of water by 0.01 mmHg, we can use Raoult's Law. This states that the vapor pressure of a solution is directly related to the mole fraction of the solvent.
Given Data
- Vapor pressure of pure water at 20°C: 17 mmHg
- Desired reduction in vapor pressure: 0.01 mmHg
- Mass of water: 1700 g
Calculating the Reduction in Vapor Pressure
- Final vapor pressure of the solution = 17 mmHg - 0.01 mmHg = 16.99 mmHg
Calculating Mole Fraction of Water
Using Raoult's Law:
- ΔP = P° * X_solute
Where:
- ΔP = change in vapor pressure
- P° = vapor pressure of pure solvent (17 mmHg)
- X_solute = mole fraction of solute (glucose)
Rearranging the formula gives us:
- X_solute = ΔP / P°
Substituting values:
- X_solute = 0.01 mmHg / 17 mmHg ≈ 0.000588
Calculating Moles of Water
- Molar mass of water = 18 g/mol
- Moles of water = 1700 g / 18 g/mol ≈ 94.44 moles
Finding Moles of Glucose
Since X_solute = moles of glucose / (moles of glucose + moles of water):
- Let moles of glucose = y
- Thus, X_solute = y / (y + 94.44)
Setting up the equation:
- 0.000588 = y / (y + 94.44)
Solving for y gives approximately:
- y ≈ 0.055 moles of glucose
Calculating Weight of Glucose
- Molar mass of glucose (C6H12O6) ≈ 180 g/mol
- Weight of glucose = 0.055 moles * 180 g/mol ≈ 9.9 g
Conclusion
To lower the vapor pressure of 1700 g of water by 0.01 mmHg at 20°C, approximately 9.9 grams of glucose should be added.
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