Calculate % of lowering of vapour pressure if 36 g glucose is added in 312g of benzene
It is assumed throughout the calculation that these two components when mixed make an ideal solution.
So from what I've learned,in an ideal solution of two components,the lowering of vapour pressure of one component equals to the mole fraction of the other component in aqueous zone.(From Raoult's law)
To get the percentage multiply it by 100℅,which will give you the required answer.
Correct me if I have mention anything which is not correct!
I believe that's the mole fraction of the other component, as stated, then times the vapor pressure of the benzene.
Percent lowering = [(vp lowering-Po)/Po]*100 = ?
To calculate the percentage of lowering of vapor pressure when 36 g of glucose is added to 312 g of benzene, we can use Raoult's law.
Raoult's law states that the vapor pressure of a solvent in a mixture is directly proportional to its mole fraction.
First, let's calculate the mole fraction of benzene and glucose in the mixture.
Moles of benzene (C6H6):
Molecular mass of benzene (C6H6) = 12 * 6 + 1 * 6 = 78 g/mol
Moles of benzene = Mass of benzene / Molecular mass of benzene = 312 g / 78 g/mol = 4 mol
Moles of glucose (C6H12O6):
Molecular mass of glucose (C6H12O6) = 12 * 6 + 1 * 12 + 16 * 6 = 180 g/mol
Moles of glucose = Mass of glucose / Molecular mass of glucose = 36 g / 180 g/mol = 0.2 mol
Now, we can calculate the mole fraction of benzene (Xbenzene) and glucose (Xglucose):
Xbenzene = Moles of benzene / Total moles of components in the mixture = 4 mol / (4 mol + 0.2 mol) = 0.9524
Xglucose = Moles of glucose / Total moles of components in the mixture = 0.2 mol / (4 mol + 0.2 mol) = 0.0476
Next, we can calculate the new vapor pressure of benzene in the mixture.
According to Raoult's law, the vapor pressure of benzene in the mixture is given by:
Pbenzene = Xbenzene * P˚benzene
where P˚benzene is the vapor pressure of pure benzene.
Assuming the vapor pressure of pure benzene is 1 atm, we have:
Pbenzene = 0.9524 * 1 atm = 0.9524 atm
The percentage of lowering of vapor pressure can be calculated as the difference between the vapor pressure of pure benzene and the vapor pressure of benzene in the mixture, divided by the vapor pressure of pure benzene, multiplied by 100.
% Lowering of vapor pressure = (P˚benzene - Pbenzene) / P˚benzene * 100
% Lowering of vapor pressure = (1 atm - 0.9524 atm) / 1 atm * 100 = 4.76 %
Therefore, the percentage of lowering of vapor pressure when 36 g of glucose is added to 312 g of benzene is approximately 4.76%.
To calculate the percentage of lowering of vapor pressure when a solute is added to a solvent, we need to use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent.
First, we need to determine the initial vapor pressure of the pure solvent (benzene). You can look up the vapor pressure of benzene at a given temperature or use experimental data.
Next, we calculate the mole fraction of the solute (glucose) in the solution.
To calculate the mole fraction (X) of glucose, use the formula:
X = moles of solute / total moles
First, we need to convert the given masses of glucose and benzene into moles using their molar masses.
The molar mass of glucose (C6H12O6) is approximately 180 g/mol. Therefore, the moles of glucose (n1) can be calculated as:
n1 = mass of glucose / molar mass of glucose
n1 = 36 g / 180 g/mol
Next, calculate the moles of benzene (n2) using its molar mass (approximately 78 g/mol):
n2 = mass of benzene / molar mass of benzene
n2 = 312 g / 78 g/mol
Now, we can calculate the mole fraction of glucose (X) using the formula mentioned earlier:
X = n1 / (n1 + n2)
Finally, we can calculate the percentage of lowering of vapor pressure using Raoult's Law. The equation is:
% lowering of vapor pressure = (1 - X) * 100
This will give you the percentage of lowering of vapor pressure when 36 g of glucose is added to 312 g of benzene.