A 298k the vapour pressure of water is 23.75mmhg.calculate the vapour pressure at same tempreature over 5% aqurous solution of urea?
23.38 mmHg
To calculate the vapor pressure of a 5% aqueous solution of urea at 298K, you can use Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
First, let's calculate the mole fraction of water in the solution. Since it is a 5% solution, we can assume that 5g of urea is dissolved in 100g of water.
Molar mass of urea (CH4N2O): 60.06 g/mol
Molar mass of water (H2O): 18.02 g/mol
Number of moles of urea:
5g / 60.06 g/mol = 0.08323 mol
Number of moles of water:
100g / 18.02 g/mol = 5.54806 mol
Total number of moles in the solution:
0.08323 mol (urea) + 5.54806 mol (water) = 5.63129 mol
Now, let's calculate the mole fraction of water:
Mole fraction of water = (moles of water) / (total moles of solution)
= 5.54806 mol / 5.63129 mol
= 0.982
Since the mole fraction of urea is (1 - mole fraction of water), the mole fraction of urea in the solution is:
Mole fraction of urea = 1 - 0.982
= 0.018
Now, using Raoult's law, we can calculate the vapor pressure of the 5% aqueous solution of urea:
Vapor pressure of solution = (mole fraction of water) x (vapor pressure of pure water)
Vapor pressure of pure water at 298K = 23.75 mmHg (given)
Vapor pressure of solution = 0.982 * 23.75 mmHg
= 23.29 mmHg
Therefore, at the same temperature of 298K, the vapor pressure of the 5% aqueous solution of urea is 23.29 mmHg.
To calculate the vapor pressure of a 5% aqueous solution of urea at the same temperature, you can use Raoult's Law. According to Raoult's Law, the vapor pressure of a component in a solution is proportional to the mole fraction of that component.
First, let's calculate the mole fraction of urea in the solution. A 5% aqueous solution means that 5g of urea is dissolved in 100g of water. The molar mass of urea is approximately 60 g/mol.
The moles of urea can be calculated as follows:
moles of urea = (mass of urea) / (molar mass of urea)
moles of urea = (5g) / (60 g/mol)
moles of urea ≈ 0.0833 mol
The moles of water can be calculated using the mass of water:
moles of water = (mass of water) / (molar mass of water)
moles of water = (100g) / (18 g/mol)
moles of water ≈ 5.56 mol
Next, calculate the mole fraction of urea:
mole fraction of urea = (moles of urea) / (moles of urea + moles of water)
mole fraction of urea = 0.0833 mol / (0.0833 mol + 5.56 mol)
mole fraction of urea ≈ 0.0148
Now, we can apply Raoult's Law to calculate the vapor pressure of the urea solution:
vapor pressure of urea solution = (vapor pressure of pure water) * (mole fraction of water)
vapor pressure of urea solution = 23.75 mmHg * 0.9852 (since mole fraction of water + mole fraction of urea = 1)
vapor pressure of urea solution ≈ 23.38 mmHg
Therefore, the vapor pressure of a 5% aqueous solution of urea, at the same temperature (298K), is approximately 23.38 mmHg.
I wonder if PCB is a subject.5% what? w/w, w/v? I'll assume 5% w/w.
That will make it 5 g urea + 95 g H2O.
mols urea = grams/molar mass = ?
mols H2O = grams/molar mass = ?
Total mols = the sume
XH2O = mols H2O/total mols.
psoln = XH2O*PoH2O
Post your work if you get stuck.