Determine the vapour pressure lowering at 25 degree Celsius of 1.25molality sucrose solution assume the solution behaves ideally (sucrose is C12H22O11). The vapour pressure of pure water at 25 degrees Celsius is 23.8torr ( hint 50.0g of sucrose was used).
delta P = XsolutePosolvent
So you must determine both X for the solute and mols for the solvent. First you must determine the grams H2O used in the experiment.
molality = 1.25 = mols sucrose/kg H2O
Solve for kg H2O. Convert to grams H2O then to mols H2O using mols = grams/molar mass.
mols sucrose = 50g/molar mass sucrose.
Total mols = mols sucrose + mols H2O
Xsucrose = mols sucrose/total mols.
Plug Xsucrose and Po into the equation at the top and solve for delta P.
Post your work if you get stuck.
Well, well, well, if it isn't Mr. Sucrose Solution! Ready to have a little fun with some calculations, are we? Alright, buckle up!
To determine the vapor pressure lowering, we need to use the formula for the vapor pressure lowering:
ΔP = i * m * Kb
Where ΔP is the vapor pressure lowering, i is the van't Hoff factor, m is the molality, and Kb is the molal boiling point constant.
Now, since sucrose doesn't ionize like a fancy electrolyte, the van't Hoff factor (i) for sucrose is 1 (because it's a single molecule). We are dealing with an ideal solution, so we can use 1 for the van't Hoff factor.
Next, we need to calculate the molality (m) of the solution, given that 50.0g of sucrose was used.
Molar mass of sucrose (C12H22O11) = 342.3 g/mol
Moles of sucrose = (50.0 g) / (342.3 g/mol)
= 0.146 mol
Now, let's calculate the molality:
Molality (m) = (moles of solute) / (mass of solvent in kg)
Given that we have 1.25 molality, we can rearrange the formula to solve for mass of solvent:
mass of solvent = (moles of solute) / (molality)
mass of solvent = (0.146 mol) / (1.25 mol/kg)
mass of solvent = 0.117 kg
Now that we have the molality and the van't Hoff factor, we can move on to calculating the vapor pressure lowering:
ΔP = (i * m * Kb)
But what's the molal boiling point constant for water, you ask? Well, for water, Kb = 0.512°C/m, or 0.512 K/kg mol.
ΔP = (1 * 1.25 mol/kg * 0.512 K/kg mol)
ΔP ≈ 0.64 K
Now, to convert the temperature change from kelvin (K) to torr, we need to use a conversion factor. For every 1 K change in boiling point, the vapor pressure decreases by approximately 0.76 torr.
So, the vapor pressure lowering is:
ΔP ≈ (0.76 torr/K * 0.64 K)
ΔP ≈ 0.49 torr
And voila! The vapor pressure lowering of your 1.25 molality sucrose solution at 25 degrees Celsius is approximately 0.49 torr. Hope this helped, Mr. Sucrose Solution! Keep the science jokes flowing like sugar! *wink*
To determine the vapor pressure lowering of a solution, we need to use the formula for the vapor pressure lowering, which is given by:
ΔP = Xsolute * Psolvent
Where ΔP is the vapor pressure lowering, Xsolute is the mole fraction of the solute, and Psolvent is the vapor pressure of the pure solvent.
First, we need to calculate the mole fraction of the solute, which is sucrose in this case. We can calculate the number of moles of sucrose by using its molar mass and the given mass of sucrose:
Molar mass of sucrose (C12H22O11) = 342.3 g/mol
Mass of sucrose = 50.0 g
Number of moles of sucrose = mass of sucrose / molar mass of sucrose
= 50.0 g / 342.3 g/mol
Next, we need to calculate the mole fraction of sucrose:
Mole fraction of sucrose = moles of sucrose / (moles of sucrose + moles of water)
Since the mass of solution is not given, we can assume a convenient mass, such as 1000 g, to calculate the mole fraction.
Moles of water = (mass of solution - moles of sucrose) / molar mass of water
= (1000 g - (50.0 g / 342.3 g/mol)) / 18.015 g/mol
Now, let's calculate the mole fraction of sucrose:
Mole fraction of sucrose = moles of sucrose / (moles of sucrose + moles of water)
Finally, we can calculate the vapor pressure lowering (ΔP):
ΔP = Xsolute * Psolvent
To determine the vapor pressure lowering of a solution, we need to apply Raoult's law. Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
The formula for calculating the vapor pressure lowering (ΔP) is:
ΔP = P°solvent - Psolution
Where:
ΔP = Vapor pressure lowering
P°solvent = Vapor pressure of the pure solvent
Psolution = Vapor pressure of the solution
First, let's calculate the mole fraction of the solvent (water) in the solution:
Mole fraction = moles of solvent / total moles of solute + solvent
We are given that 50.0g of sucrose (C12H22O11) was used, which has a molar mass of 342.3 g/mol. So, the number of moles of sucrose is:
Moles of sucrose = mass of sucrose / molar mass of sucrose
Moles of sucrose = 50.0g / 342.3 g/mol
Next, we can calculate the number of moles of water (solvent) using the fact that the solution is 1.25 molality, which means there are 1.25 moles of sucrose for every kilogram of water (solvent). Since we know that 1 kg of water is equal to 1000g, we can calculate the moles of water:
Moles of water = molality * Mass of water
Moles of water = 1.25 mol/kg * 1000 g
Now, we can calculate the mole fraction of water:
Mole fraction of water = moles of water / (moles of sucrose + moles of water)
After calculating the mole fraction of water, we use Raoult's law to calculate the vapor pressure of the sucrose solution:
Psolution = P°solvent * Mole fraction of water
Finally, we can calculate the vapor pressure lowering:
ΔP = P°solvent - Psolution
Substituting the given value of P°solvent (23.8 torr) and solving the equation, we can determine the vapor pressure lowering at 25 degrees Celsius of the 1.25 molality sucrose solution.